Question 2

   Question: $AB$ is a part of an electrical circuit (see figure). The potential difference "$V_A - V_B$", at the instant when current $i = 2$ A and is increasing at a rate of 1 amp/second is:

   Options: 

       A. 9 volt

       B. 10 volt

       C. 5 volt

       D. 6 volt

   Correct Answer: B

   Year: NEET 2025

   Solution (as Given in the Source):     Given, $i = 2$ A and $\frac{di}{dt} = +1 \text{ A/s}$

    $V_A - L\frac{di}{dt} - 5 - i \times 2 = V_B$

    $\Rightarrow V_A - 1 \times 1 - 5 - 2 \times 2 = V_B$

    $\Rightarrow V_A - V_B = 10 \text{ volt}$

   Step Solution:

    1.  Identify given values: $i = 2$ A, $\frac{di}{dt} = 1 \text{ A/s}$, $E = 5$ V, $R = 2 \Omega$, and $L = 1$ H (inferred from the calculation $L \cdot \frac{di}{dt} = 1 \times 1$).

    2.  Apply Kirchhoff’s Voltage Law (KVL): Start from point A and move toward B, subtracting voltage drops across components: $V_A - V_L - V_E - V_R = V_B$.

    3.  Substitute component formulas: $V_A - (L \frac{di}{dt}) - 5 - (i \times R) = V_B$.

    4.  Insert numerical values: $V_A - (1 \times 1) - 5 - (2 \times 2) = V_B$.

    5.  Solve for $V_A - V_B$: $V_A - 1 - 5 - 4 = V_B \Rightarrow V_A - 10 = V_B \Rightarrow V_A - V_B = 10$ V.

   Difficulty Level: Medium

   Concept Name: Kirchhoff’s Voltage Law (KVL) and Self-Inductance.

   Short cut solution: Direct summation of voltage drops in the direction of current: $\Delta V = (L \cdot \frac{di}{dt}) + E_{battery} + (i \cdot R) = (1 \cdot 1) + 5 + (2 \cdot 2) = 10$ V.

Question 9

   Question: The net magnetic flux through any closed surface is

   Options: 

       A. Positive

       B. Infinity

       C. Negative

       D. Zero

   Correct Answer: D

   Year: NEET 2023

   Solution (as Given in the Source): Magnetic monopole doesn't exist. Hence net magnetic flux through any closed surface is zero.

   Step Solution:

    1.  Recall Gauss's Law for Magnetism: The total magnetic flux through any closed surface is always zero ($\oint \mathbf{B} \cdot d\mathbf{A} = 0$).

    2.  Identify the reason: Magnetic field lines form continuous loops; there are no isolated magnetic charges (monopoles).

    3.  Analyze flux entry and exit: Every magnetic field line that enters a closed surface must also exit it.

    4.  Conclude: Since the lines entering equal the lines leaving, the net flux is zero.

   Difficulty Level: Easy

   Concept Name: Gauss's Law for Magnetism.

   Short cut solution: Use the fundamental property that magnetic field lines are always closed loops, making the net flux through any closed surface zero by definition.

Question 17

   Question: An emf is generated by an ac generator having 100 turn coil, of loop area 1 m$^2$. The coil rotates at a speed of one revolution per second and placed in a uniform magnetic field of 0.05 T perpendicular to the axis of rotation of the coil. The maximum value of emf is :-

   Options: 

       A. 3.14 V

       B. 31.4 V

       C. 62.8 V

       D. 6.28 V

   Correct Answer: B

   Year: NEET 2023 mpr

   Solution (as Given in the Source): 

    $\omega = 2\pi \frac{\text{rad}}{\text{sec}}$

    $E_{\max} = NBA\omega$

    $= 100 \times 0.05 \times 1 \times 2\pi$

    $= 10 \times \pi$

    $= 31.4 \text{ V}$

   Step Solution:

    1.  List parameters: Number of turns $N = 100$, Area $A = 1 \text{ m}^2$, Magnetic field $B = 0.05$ T, Frequency $f = 1$ rev/s.

    2.  Calculate angular frequency ($\omega$): $\omega = 2\pi f = 2 \times \pi \times 1 = 2\pi$ rad/s.

    3.  Apply maximum EMF formula: $E_{\max} = NBA\omega$.

    4.  Substitute values: $E_{\max} = 100 \times 0.05 \times 1 \times 2\pi$.

    5.  Final Calculation: $E_{\max} = 5 \times 2\pi = 10\pi = 10 \times 3.14 = 31.4$ V.

   Difficulty Level: Easy

   Concept Name: Maximum Induced EMF in an AC Generator.

   Short cut solution: Quickly multiply the integers: $100 \times 0.05 = 5$. Since $\omega = 2\pi$, $E_{\max} = 5 \times 2\pi = 10\pi \approx 31.4$ V.

Question 19

   Question: The magnetic flux linked to a circular coil of radius R is: $\Phi = 2t^3 + 4t^2 + 2t + 5$ Wb. The magnitude of induced emf in the coil at $t = 5$ s is:

   Options: 

       A. 192 V

       B. 108 V

       C. 197 V

       D. 150 V

   Correct Answer: A

   Year: NEET Re-2022

   Solution (as Given in the Source): 

    $\varphi = 2t^3 + 4t^2 + 2t + 5$

    $|e| = \left| \frac{d\varphi}{dt} \right| = \frac{d}{dt} [2t^3 + 4t^2 + 2t + 5] = 6t^2 + 8t + 2$

    $e(t = 5) = 6(5)^2 + 8(5) + 2 = 150 + 40 + 2 = 192 \text{ V}$

   Step Solution:

    1.  State the Flux equation: $\Phi = 2t^3 + 4t^2 + 2t + 5$.

    2.  Apply Faraday's Law: The induced EMF is the time derivative of flux, $e = \frac{d\Phi}{dt}$.

    3.  Differentiate the expression: $\frac{d}{dt}(2t^3 + 4t^2 + 2t + 5) = (3 \times 2)t^2 + (2 \times 4)t + 2 = 6t^2 + 8t + 2$.

    4.  Substitute $t = 5$: $e = 6(5)^2 + 8(5) + 2$.

    5.  Calculate final value: $e = 150 + 40 + 2 = 192 \text{ V}$.

   Difficulty Level: Easy

   Concept Name: Faraday’s Law of Induction and Power Rule of Differentiation.

   Short cut solution: Directly differentiate and plug in 5: $6(25) + 40 + 2 = 192 \text{ V}$.

Question 24

   Question: A square loop of side 1m and resistance 1$\Omega$ is placed in a magnetic field of 0.5 T. If the plane of loop is perpendicular to the direction of magnetic field, the magnetic flux through the loop is

   Options: 

       A. 2 weber

       B. 0.5 weber

       C. 1 weber

       D. Zero weber

   Correct Answer: B

   Year: NEET-2022

   Solution (as Given in the Source):

Magnetic flux $(\varphi_B) = \vec{B} \cdot \vec{A}$. $B$ and $A$ are in same direction, therefore $\varphi_B = B \cdot A = 0.5 \times 1^2 = 0.5 \text{ Wb}$.

   Step Solution:

    1.  Find the Area (A): For a square, $Area = side^2 = 1 \times 1 = 1 \text{ m}^2$.

    2.  Determine the Angle ($\theta$): Since the plane is perpendicular to the field, the area vector (normal to the plane) is parallel to the field, so $\theta = 0^\circ$.

    3.  Use the Flux Formula: $\Phi = BA \cos \theta$.

    4.  Substitute values: $\Phi = (0.5 \text{ T}) \times (1 \text{ m}^2) \times \cos 0^\circ$.

    5.  Solve: $\Phi = 0.5 \times 1 \times 1 = 0.5 \text{ Weber}$.

   Difficulty Level: Easy

   Concept Name: Magnetic Flux Calculation.

   Short cut solution: When the plane is perpendicular to the field, flux is simply $B \times A$. Here, $0.5 \times 1 = 0.5 \text{ Wb}$.

Question 26

   Question: A big circular coil of 1000 turns and average radius 10m is rotating about its horizontal diameter at 2 rad s⁻¹. If the vertical component of earth's magnetic field at that place is $2 \times 10^{-5}$ T and electrical resistance of the coil is 12.56$\Omega$, then the maximum induced current in the coil will be

   Options: 

       A. 0.25 A

       B. 1.5 A

       C. 1 A

       D. 2 A

   Correct Answer: C

   Year: NEET-2022

   Solution (as Given in the Source): 

    $\varphi_B = NBA \cos \omega t$

    $\varepsilon = \frac{-d\varphi_B}{dt} = NBA\omega \sin \omega t$

    $i_{\max} = \frac{\varepsilon_{\max}}{R} = \frac{NBA\omega}{R} = \frac{1000 \times 2 \times 10^{-5} \times \pi(10)^2 \times 2}{12.56} = 1 \text{ A}$

   Step Solution:

    1.  Identify parameters: $N = 1000$, $B = 2 \times 10^{-5}$ T, $r = 10$ m, $\omega = 2$ rad/s, $R = 12.56 \Omega$.

    2.  Calculate Area (A): $A = \pi r^2 = \pi (10)^2 = 100\pi \text{ m}^2$.

    3.  Find Max EMF ($\varepsilon_{\max}$): $\varepsilon_{\max} = NBA\omega = 1000 \times (2 \times 10^{-5}) \times 100\pi \times 2$.

    4.  Simplify EMF: $\varepsilon_{\max} = 1000 \times 100 \times 2 \times 2 \times 10^{-5} \times \pi = 4\pi \text{ V}$.

    5.  Calculate Max Current ($i_{\max}$): $i_{\max} = \frac{\varepsilon_{\max}}{R} = \frac{4\pi}{12.56}$. Since $\pi \approx 3.14$, $4\pi \approx 12.56$, therefore $i_{\max} = \frac{12.56}{12.56} = 1 \text{ A}$.

   Difficulty Level: Medium

   Concept Name: Induced EMF in a Rotating Coil (AC Generator).

   Short cut solution: Use $i_{\max} = \frac{NBA\omega}{R}$. Notice that $4\pi \approx 12.56$, so if the numerator has $4\pi$ and the denominator is $12.56$, they cancel out to leave $1$.

Question 34

   Question: A 800 turn coil of effective area $0.05 \text{ m}^2$ is kept perpendicular to a magnetic field $5 \times 10^{-5}$ T. When the plane of the coil is rotated by $90^\circ$ around any of its coplanar axis in 0.1 s, the emf induced in the coil will be

   Options: 

       A. 0.02V

       B. 2V

       C. 0.2V

       D. $2 \times 10^{-3}$ V

   Correct Answer: A

   Year: NEET 2019

   Solution (as Given in the Source): 

    $\text{Here } N = 800, A = 0.05 \text{ m}^2, B = 5 \times 10^{-5} \text{ T}, \Delta t = 0.1 \text{ s}$

    $\text{Induced emf}, \varepsilon = -\frac{\Delta \phi}{\Delta t} = -\frac{(\phi_f - \phi_i)}{\Delta t}$

    $\phi_i = N(\vec{B} \cdot \vec{A}) = 800 \times 5 \times 10^{-5} \times 0.05 \times \cos 0^\circ = 2 \times 10^{-3} \text{ T m}^2$

    $\phi_f = 0$

    $\therefore \varepsilon = \frac{-(0 - 2 \times 10^{-3})}{0.1} = 2 \times 10^{-2} \text{ V} = 0.02 \text{ V}$

   Step Solution:

    1.  Calculate Initial Flux ($\phi_i$): Since the field is perpendicular to the plane, $\theta = 0^\circ$; $\phi_i = NBA \cos 0^\circ = 800 \times (5 \times 10^{-5}) \times 0.05 = 2 \times 10^{-3} \text{ Wb}$.

    2.  Calculate Final Flux ($\phi_f$): Rotating by $90^\circ$ makes the plane parallel to the field ($\theta = 90^\circ$); $\phi_f = NBA \cos 90^\circ = 0$.

    3.  Find Change in Flux: $\Delta \phi = \phi_f - \phi_i = 0 - 2 \times 10^{-3} = -2 \times 10^{-3} \text{ Wb}$.

    4.  Apply Faraday’s Law: Induced emf $\varepsilon = -\frac{\Delta \phi}{\Delta t}$.

    5.  Final Calculation: $\varepsilon = -\frac{(-2 \times 10^{-3})}{0.1} = \frac{0.002}{0.1} = 0.02 \text{ V}$.

   Difficulty Level: Medium

   Concept Name: Faraday’s Law of Induction and Magnetic Flux.

   Short cut solution: $\varepsilon = \frac{NBA}{\Delta t} = \frac{800 \times 5 \times 10^{-5} \times 0.05}{0.1} = \frac{0.002}{0.1} = 0.02 \text{ V}$.

Question 36

   Question: A cycle wheel of radius 0.5m is rotated with constant angular velocity of 10rad/s in a region of magnetic field of 0.1T which is perpendicular to the plane of the wheel. The EMF generated between its centre and the rim is,

   Options: 

       A. 0.25V

       B. 0.125V

       C. 0.5V

       D. zero

   Correct Answer: B

   Year: OD NEET 2019

   Solution (as Given in the Source): 

    Here, given, $B = 0.1 \text{ T}, r = 0.5 \text{ m}, \omega = 10 \text{ rad/s}$.

    So, the emf generated between its centre and rim is, $\varepsilon = \frac{1}{2} B \omega r^2 = \frac{1}{2} \times 0.1 \times 10 \times (0.5)^2 = 0.125 \text{ V}$

   Step Solution:

    1.  Identify Parameters: $B = 0.1 \text{ T}$, radius $r = 0.5 \text{ m}$, and $\omega = 10 \text{ rad/s}$.

    2.  State Motional EMF Formula: For a rotating conductor in a magnetic field, $\varepsilon = \frac{1}{2} B \omega r^2$.

    3.  Substitute Values: $\varepsilon = \frac{1}{2} \times 0.1 \times 10 \times (0.5)^2$.

    4.  Simplify Constants: $\varepsilon = 0.5 \times (0.25)$.

    5.  Calculate Final Value: $\varepsilon = 0.125 \text{ V}$.

   Difficulty Level: Easy

   Concept Name: Motional EMF in a Rotating Conductor.

   Short cut solution: Use the standard formula $\varepsilon = \frac{1}{2} B \omega r^2$ directly: $0.5 \times 0.1 \times 10 \times 0.25 = 0.125 \text{ V}$.

Question 45

   Question: A uniform magnetic field is restricted within a region of radius r. The magnetic field changes with time at a rate $\frac{dB}{dt}$. Loop 1 of radius $R > r$ encloses the region r and loop 2 of radius R is outside the region of magnetic field as shown in the figure. Then the e.m.f. generated is

   Options: 

       A. zero in loop 1 and zero in loop 2

       B. $-\frac{dB}{dt} \pi r^2$ in loop 1 and $-\frac{dB}{dt} \pi r^2$ in loop 2

       C. $\frac{dB}{dt} \pi R^2$ in loop 1 and zero in loop 2

       D. $\frac{dB}{dt} \pi r^2$ in loop 1 and zero in loop 2

   Correct Answer: D

   Year: 2016 NEET Phase-II

   Solution (as Given in the Source): 

    Emf generated in loop 1, $\varepsilon_1 = -\frac{d\phi}{dt} = -\frac{d}{dt} (\vec{B} \cdot \vec{A}) = -\frac{d}{dt} (BA) = -A \times \frac{dB}{dt} = -(\pi r^2 \frac{dB}{dt})$

    ($\because A = \pi r^2$ because $\frac{dB}{dt}$ is restricted up to radius r).

    Emf generated in loop 2, $\phi_2 = -\frac{d}{dt} (BA) = -\frac{d}{dt} (0 \times A) = 0$

   Step Solution:

    1.  Define Flux for Loop 1: The loop encloses the entire magnetic area $\pi r^2$, so $\phi_1 = B \times (\pi r^2)$.

    2.  Apply Faraday's Law for Loop 1: $\varepsilon_1 = -\frac{d\phi_1}{dt} = -\frac{d(B \pi r^2)}{dt}$.

    3.  Differentiate for Loop 1: Since $r$ is constant, $\varepsilon_1 = -\pi r^2 \frac{dB}{dt}$. (Magnitude is $\frac{dB}{dt} \pi r^2$).

    4.  Analyze Loop 2: Loop 2 is entirely outside the region where $B$ exists, so the magnetic field $B$ within its area is zero.

    5.  Determine EMF for Loop 2: Since flux $\phi_2 = 0$, its derivative $\frac{d\phi_2}{dt}$ is also 0; therefore, $\varepsilon_2 = 0$.

   Difficulty Level: Medium

   Concept Name: Faraday’s Law of Induction and Effective Area of Magnetic Flux.

   Short cut solution: Emf is produced only where flux changes. Loop 2 has no flux (0). Loop 1 encloses a field changing over area $\pi r^2$, so its emf is $\frac{dB}{dt} \pi r^2$.

Question 48

   Question: A conducting square frame of side ‘a’ and a long straight wire carrying current I are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity ‘V’. The emf induced in the frame will be proportional to:

   Options: 

       A. $\frac{1}{(2x+a)^2}$

       B. $\frac{1}{(2x-a)(2x+a)}$

       C. $\frac{1}{x^2}$

       D. $\frac{1}{(2x-a)^2}$

   Correct Answer: B

   Year: 2015

   Solution (as Given in the Source): 

    $= \frac{\mu_0 I}{2\pi(x - a/2)}aV - \frac{\mu_0 I}{2\pi(x + a/2)}aV$

    $= \frac{\mu_0 I}{2\pi} \left[ \frac{1}{(2x-a)} - \frac{2}{(2x+a)} \right] aV$ (Note: source contains a minor typo in the coefficients)

    $= \frac{\mu_0 I}{2\pi} \times 2 \left[ \frac{2a}{(2x-a)(2x+a)} \right]$

    $\therefore \varepsilon \propto \frac{1}{(2x-a)(2x+a)}$

   Step Solution:

    1.  Identify distances: The two vertical sides of the frame are at distances $r_1 = x - \frac{a}{2}$ and $r_2 = x + \frac{a}{2}$ from the wire.

    2.  Determine magnetic fields: Using $B = \frac{\mu_0 I}{2\pi r}$, find $B_1 = \frac{\mu_0 I}{2\pi(x - a/2)}$ and $B_2 = \frac{\mu_0 I}{2\pi(x + a/2)}$.

    3.  Apply Motional EMF ($\varepsilon = Blv$): The net EMF is the difference between EMFs of the two sides: $\varepsilon = (B_1 - B_2)aV$.

    4.  Substitute and simplify: $\varepsilon = \frac{\mu_0 I a V}{2\pi} \left( \frac{1}{x - a/2} - \frac{1}{x + a/2} \right) = \frac{\mu_0 I a V}{2\pi} \left( \frac{2}{2x - a} - \frac{2}{2x + a} \right)$.

    5.  Final Proportionality: $\varepsilon = \frac{\mu_0 I a V}{\pi} \left( \frac{(2x+a) - (2x-a)}{(2x-a)(2x+a)} \right) = \frac{2 \mu_0 I a^2 V}{\pi (2x-a)(2x+a)}$. Thus, $\varepsilon \propto \frac{1}{(2x-a)(2x+a)}$.

   Difficulty Level: Hard

   Concept Name: Motional EMF and Ampere’s Law.

   Short cut solution: The EMF is proportional to the difference in magnetic field strengths at the two sides: $\varepsilon \propto (\frac{1}{r_1} - \frac{1}{r_2}) = \frac{r_2 - r_1}{r_1 r_2}$. Since $r_2 - r_1 = a$ (constant), $\varepsilon \propto \frac{1}{(x-a/2)(x+a/2)}$, which simplifies to $\frac{1}{(2x-a)(2x+a)}$.

Question 50

   Question: An electron moves on a straight line path XY as shown. The abcd is a coil adjacent to the path of electron. What will be the direction of current, if any, induced in the coil?

   Options: 

       A. The current will reverse its direction as the electron goes past the coil

       B. No current induced

       C. abcd

       D. adcb

   Correct Answer: A

   Year: 2015

   Solution (as Given in the Source):

When the electron moves from X to Y, the flux linked with the coil abcd (which is into the page) will first increase and then decrease as the electron passes by. So the induced current in the coil will be first anticlockwise and will reverse its direction (i.e. will become clockwise) as the electron goes past the coil.

   Step Solution:

    1.  Determine field direction: A moving electron (negative charge) creates a magnetic field into the page in the region of the coil.

    2.  Analyze approach phase: As the electron approaches, the magnetic field strength increases, causing the flux into the page to increase.

    3.  Apply Lenz's Law (Initial): To oppose the increase, the induced field must be out of the page, resulting in an anticlockwise current.

    4.  Analyze receding phase: As the electron moves away, the field strength decreases, causing the flux into the page to decrease.

    5.  Apply Lenz's Law (Final): To oppose the decrease, the induced field must be into the page, resulting in a clockwise current.

   Difficulty Level: Medium

   Concept Name: Lenz’s Law and Magnetic Field of a Moving Charge.

   Short cut solution: The flux linked with the loop increases as the charge approaches and decreases as it leaves; any such change in flux gradient necessarily causes the induced EMF to change sign, reversing the current.

Question 51

   Question: A thin semicircular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in the figure. The potential difference developed across the ring when its speed is v, is:

   Options: 

       A. zero

       B. $\frac{Bv\pi r^2}{2}$ and P is at higher potential

       C. $\pi rBv$ and R is at higher potential

       D. 2rBv and R is at higher potential

   Correct Answer: D

   Year: 2014

   Solution (as Given in the Source): Motional emf induced in the semicircular ring PQR is equivalent to the motional emf induced in the imaginary conductor PR. $\varepsilon_{PQR} = \varepsilon_{PR} = Bvl = Bv(2r)$ ($l = PR = 2r$). Therefore, potential difference developed across the ring is 2rBv with R is at higher potential.

   Step Solution:

    1.  Identify effective length: For motional EMF in a curved wire, the effective length $l$ is the straight-line distance between the endpoints perpendicular to the motion.

    2.  Calculate $l$: The distance between endpoints P and R of the semicircle is the diameter, $l = 2r$.

    3.  State formula: The induced motional EMF is given by $\varepsilon = Bvl$.

    4.  Substitute values: $\varepsilon = B \times v \times (2r) = 2rBv$.

    5.  Determine Polarity: Using the right-hand rule ($\vec{v} \times \vec{B}$), the force on positive charges is directed toward point R, making R at a higher potential.

   Difficulty Level: Easy

   Concept Name: Motional EMF in a Curved Conductor.

   Short cut solution: Always replace a curved conductor with a straight line connecting its ends. The distance is $2r$, so the EMF is simply $B \times v \times 2r$.

Question 53

   Question: A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced e.m.f. is

   Options: 

       A. four times per revolution

       B. six times per revolution

       C. once per revolution

       D. twice per revolution

   Correct Answer: D

   Year: 2013 NEET

   Solution (as Given in the Source): (The source provides the answer but the detailed explanation text is omitted in the excerpt).

   Step Solution:

    1.  Identify the process: A loop rotating in a magnetic field acts as an AC generator where induced EMF is $e = NBA\omega \sin(\omega t)$.

    2.  Analyze the cycle: In one full revolution ($360^\circ$ or $2\pi$ radians), the sine function goes through one complete cycle.

    3.  Determine sign changes: A sine wave has one positive half-cycle and one negative half-cycle.

    4.  Count direction changes: The direction of the induced EMF reverses when the sine function crosses zero. 

    5.  Conclusion: This occurs twice in one revolution (at $180^\circ$ and $360^\circ$), so the direction changes twice per revolution.

   Difficulty Level: Easy

   Concept Name: AC Generator Induced EMF.

   Short cut solution: One full rotation in an AC generator corresponds to one full sine wave; a sine wave changes polarity twice per period.

Question 56

   Question: The current ($I$) in the inductance is varying with time according to the plot shown in figure. Which one of the following is the correct variation of voltage with time in the coil?

   Options:

  A. 

  B. 

  C. 

  D. 

   Correct Answer: D

   Year: 2012

   Solution (as Given in the Source): $V = -L \frac{dI}{dt}$; $V \propto$ slope of $I - t$ graph.

   Step Solution:

    1.  State the relationship: The induced voltage across an inductor is given by $V = -L \frac{dI}{dt}$.

    2.  Analyze segment 1: In the first half of the graph, current $I$ increases linearly with time, so the slope $\frac{dI}{dt}$ is a positive constant.

    3.  Determine initial Voltage: Since $V = -L \times (\text{positive constant})$, the voltage is a negative constant.

    4.  Analyze segment 2: In the second half, current $I$ decreases linearly, making the slope $\frac{dI}{dt}$ a negative constant.

    5.  Determine final Voltage: Since $V = -L \times (\text{negative constant})$, the voltage becomes a positive constant.

   Difficulty Level: Medium

   Concept Name: Lenz's Law and Inductor Voltage-Current Relationship.

   Short cut solution: Voltage is the negative slope of the $I-t$ graph. A constant positive slope gives a negative constant voltage, and a constant negative slope gives a positive constant voltage.

Question 60

   Question: In a coil of resistance 10 $\Omega$, the induced current developed by changing magnetic flux through it, is shown in figure as a function of time. The magnitude of change in flux through the coil in weber is

   Options: 

       A. 8

       B. 2

       C. 6

       D. 4

   Correct Answer: B

   Year: 2012 Mains

   Solution (as Given in the Source): $q = \text{Area} = \frac{1}{2} \times 4 \times 0.1 = 0.2 \text{ C}$. As $q = \frac{\Delta \phi}{R}$, $\therefore \Delta \phi = qR = (0.2 \text{ C}) (10 \Omega) = 2 \text{ weber}$.

   Step Solution:

    1.  Calculate Charge ($q$): The total charge passed is the area under the current-time ($i-t$) graph: $q = \int i dt$.

    2.  Determine Area: For the given triangular graph, $q = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 0.1 \text{ s} \times 4 \text{ A} = 0.2 \text{ C}$.

    3.  Relate Charge to Flux: Use the formula $q = \frac{\Delta \phi}{R}$, where $\Delta \phi$ is the change in flux and $R$ is resistance.

    4.  Isolate $\Delta \phi$: $\Delta \phi = q \times R$.

    5.  Substitute values: $\Delta \phi = 0.2 \text{ C} \times 10 \Omega = 2$ weber.

   Difficulty Level: Medium

   Concept Name: Relationship between Induced Charge and Flux Change.

   Short cut solution: Change in flux is the resistance multiplied by the area under the $i-t$ curve: $\Delta \phi = 10 \times (\frac{1}{2} \times 4 \times 0.1) = 10 \times 0.2 = 2 \text{ Wb}$.

Question 63

   Question: The current $i$ in a coil varies with time as shown in the figure. The variation of induced emf with time would be:

   Options: 

       A.

       B.

       C.

       D.

   Correct Answer: A

   Year: 2011

   Solution (as Given in the Source): (The source indicates the answer is A but the explicit text solution for this specific question is not detailed; however, it follows the principle $V = -L \frac{di}{dt}$ used in Question 56).

   Step Solution:

    1.  State the Relationship: The induced EMF is given by $e = -L \frac{di}{dt}$, meaning it is proportional to the negative slope of the $i-t$ graph.

    2.  Analyze Phase 1 (Linear Increase): The current increases at a constant rate, so $\frac{di}{dt}$ is a positive constant. Therefore, $e$ is a negative constant.

    3.  Analyze Phase 2 (Constant Current): The current is horizontal/constant, so $\frac{di}{dt} = 0$. Therefore, $e$ is zero.

    4.  Analyze Phase 3 (Linear Decrease): The current decreases at a constant rate, so $\frac{di}{dt}$ is a negative constant. Therefore, $e$ is a positive constant.

    5.  Match Graph: Graph A correctly displays these three distinct constant levels (negative, zero, positive).

   Difficulty Level: Medium

   Concept Name: Lenz's Law and Faraday’s Law of Induction.

   Short cut solution: EMF is the negative derivative of current; look for the graph that represents the negative slope of each segment of the $i-t$ plot.

Question 66

   Question: A conducting circular loop is placed in a uniform magnetic field, $B = 0.025$ T with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of 1 mm/s. The induced emf when the radius is 2 cm, is:

   Options: 

       A. $2\pi \mu\text{V}$

       B. $\pi \mu\text{V}$

       C. $\pi/2 \mu\text{V}$

       D. $2 \mu\text{V}$

   Correct Answer: B

   Year: 2010

   Solution (as Given in the Source): 

    $|\varepsilon| = \frac{d\phi}{dt} = \frac{d}{dt} (B \pi r^2) = B \pi (2r) \frac{dr}{dt}$

    $= 0.025 \times \pi \times 2 \times (2 \times 10^{-2}) \times (1 \times 10^{-3})$

    $= \pi \times 10^{-6} \text{ V} = \pi \mu\text{V}$.

   Step Solution:

    1.  Identify parameters: $B = 0.025$ T, $r = 0.02$ m, and the rate of change $\frac{dr}{dt} = 1 \times 10^{-3}$ m/s.

    2.  Define Flux ($\phi$): Since the plane is perpendicular to the field, $\phi = B \cdot A = B(\pi r^2)$.

    3.  Apply Faraday's Law: $\varepsilon = \frac{d\phi}{dt}$. Using the chain rule, $\frac{d}{dt}(B \pi r^2) = B \pi (2r \frac{dr}{dt})$.

    4.  Substitute Values: $|\varepsilon| = 0.025 \times \pi \times 2 \times 0.02 \times 0.001$.

    5.  Calculate: $|\varepsilon| = 0.05 \times \pi \times 0.02 \times 0.001 = 0.001 \times \pi \times 0.001 = 10^{-6}\pi \text{ V} = \pi \mu\text{V}$.

   Difficulty Level: Medium

   Concept Name: Faraday’s Law and the Chain Rule of Differentiation.

   Short cut solution: Use the formula $|\varepsilon| = 2 \pi B r (\frac{dr}{dt})$: $2 \times \pi \times 0.025 \times 0.02 \times 0.001 = \pi \mu\text{V}$.

Question 70

   Question: A rectangular, a square, a circular and an elliptical loop, all in the (x-y) plane, are moving out of a uniform magnetic field with a constant velocity, $\vec{v} = v\hat{i}$. The magnetic field is directed along the negative z-axis direction. The induced emf, during the passage of these loops out of the field region, will not remain constant for:

   Options: 

       A. the circular and the elliptical loops

       B. only the elliptical loop

       C. any of the four loops

       D. the rectangular, circular and elliptical loops

   Correct Answer: A

   Year: 2009

   Solution (as Given in the Source): Once a rectangular loop or a square loop is being drawn out of the field, the rate of cutting the lines of field will be a constant for a square and rectangle, but not for circular or elliptical areas.

   Step Solution:

    1.  Understand Induced EMF: $\varepsilon = Blv$, where $l$ is the "effective length" of the conductor cutting the field lines perpendicular to the motion.

    2.  Analyze Rectangular/Square Loops: As they move out, the length of the side perpendicular to the velocity is constant. Thus, $l$ is constant and $\varepsilon$ is constant.

    3.  Analyze Circular/Elliptical Loops: As they move out, the "vertical width" of the part remaining in the field changes continuously because of the curved boundary.

    4.  Relate to Area Change: The rate of change of area ($\frac{dA}{dt}$) for a circle or ellipse moving at constant velocity is not constant.

    5.  Conclusion: Since $\varepsilon = B \frac{dA}{dt}$, the EMF will vary for circular and elliptical loops.

   Difficulty Level: Medium

   Concept Name: Motional EMF and Faraday’s Law of Induction.

   Short cut solution: EMF is constant only if the width of the loop perpendicular to its motion is constant. Circular and elliptical loops have changing widths, so their EMF is not constant.

Question 72

   Question: A conducting circular loop is placed in a uniform magnetic field 0.04 T with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at 2 mm/s. The induced emf in the loop when the radius is 2 cm is

   Options: 

       A. 4.8π µV

       B. 0.8π µV

       C. 1.6π µV

       D. 3.2π µV

   Correct Answer: D

   Year: 2009

   Solution (as Given in the Source): 

    $\phi = BA \cos \theta = B (\pi r^2) \cos 0^\circ = B \pi r^2$

    $| \varepsilon | = \frac{d\phi}{dt} = B \pi (2r) \frac{dr}{dt}$

    $= 0.04 \times \pi \times (2 \times 2 \times 10^{-2}) \times 2 \times 10^{-3} = 3.2 \pi \times 10^{-6} \text{ V} = 3.2 \pi \mu\text{V}$

   Step Solution:

    1.  Identify parameters: $B = 0.04$ T, $r = 0.02$ m, and the rate of change $\frac{dr}{dt} = 2 \times 10^{-3}$ m/s.

    2.  State the Flux ($\phi$): Since the plane is perpendicular to the field, the angle between the area vector and $B$ is $0^\circ$, so $\phi = B(\pi r^2)$.

    3.  Apply Faraday's Law: EMF $|\varepsilon| = \frac{d\phi}{dt} = \frac{d}{dt}(B \pi r^2)$.

    4.  Use the Chain Rule: $|\varepsilon| = B \pi (2r \frac{dr}{dt})$.

    5.  Calculate final value: $|\varepsilon| = 0.04 \times \pi \times (2 \times 0.02) \times 0.002 = 3.2\pi \times 10^{-6} \text{ V} = 3.2\pi \mu\text{V}$.

   Difficulty Level: Medium

   Concept Name: Faraday’s Law of Induction and Motional EMF in a Shrinking Loop.

   Short cut solution: Use the formula $|\varepsilon| = 2 \pi B r (\frac{dr}{dt})$ directly: $2 \times \pi \times 0.04 \times 0.02 \times 0.002 = 3.2\pi \mu\text{V}$.

Question 75

   Question: A circular disc of radius 0.2 meter is placed in a uniform magnetic field of induction $\frac{1}{\pi} (\frac{\text{Wb}}{\text{m}^2})$ in such a way that its axis makes an angle of $60^\circ$ with $\vec{B}$. The magnetic flux linked with the disc is

   Options: 

       A. 0.08 Wb

       B. 0.01 Wb

       C. 0.02 Wb

       D. 0.06 Wb

   Correct Answer: C

   Year: 2008

   Solution (as Given in the Source): 

    $B = \frac{1}{\pi} (\frac{\text{Wb}}{\text{m}^2})$

    Area of the disc normal to B is $\pi R^2 \cos 60^\circ$

    Flux = B x Area normal

    $\therefore \text{Flux} = \frac{1}{2} \times 0.04 = 0.02 \text{ Wb}$

   Step Solution:

    1.  Identify parameters: $B = \frac{1}{\pi}$ T, $r = 0.2$ m, and the angle $\theta = 60^\circ$ (angle between the axis/normal and the field).

    2.  Calculate Area ($A$): $A = \pi r^2 = \pi (0.2)^2 = 0.04\pi \text{ m}^2$.

    3.  State the Flux formula: $\Phi = BA \cos \theta$.

    4.  Substitute values: $\Phi = (\frac{1}{\pi}) \times (0.04\pi) \times \cos 60^\circ$.

    5.  Solve: $\Phi = 0.04 \times 0.5 = 0.02 \text{ Wb}$.

   Difficulty Level: Easy

   Concept Name: Magnetic Flux Definition.

   Short cut solution: Notice $\pi$ in the area formula cancels the $\pi$ in the denominator of $B$. Flux simply becomes $r^2 \cos \theta = (0.2)^2 \times 0.5 = 0.04 \times 0.5 = 0.02$ Wb.

Question 84

   Question: As a result of change in the magnetic flux linked to the closed loop as shown in the figure, an e.m.f. V volt is induced in the loop. The work done (joule) in taking a charge Q coulomb once along the loop is

   Options: 

       A. QV

       B. 2QV

       C. QV/2

       D. zero

   Correct Answer: A

   Year: 2005

   Solution (as Given in the Source): Work done due to a charge $W = QV$

   Step Solution:

    1.  Understand EMF definition: Electromotive force (EMF) is defined as the work done per unit charge in moving a charge once around a complete circuit.

    2.  State the formula: $V = \frac{W}{Q}$.

    3.  Isolate Work ($W$): $W = Q \times V$.

    4.  Analyze the query: The charge $Q$ is taken "once along the loop" with an induced EMF of $V$.

    5.  Finalize: The work done is the product of charge and potential difference (EMF), which is $QV$.

   Difficulty Level: Easy

   Concept Name: Definition of Induced EMF and Work Done.

   Short cut solution: By the fundamental definition of potential difference/EMF, Work = Charge $\times$ Potential. Thus, $W = QV$.

Question 88

   Question: For a coil having L = 2 mH, current flow through it is $I = t^2 e^{-t}$ then, the time at which emf becomes zero.

   Options: 

       A. 2 sec

       B. 1 sec

       C. 4 sec

       D. 3 sec

   Correct Answer: A

   Year: 2001

   Solution (as Given in the Source): $I = t^2 e^{-t}$; $|\varepsilon| = L \frac{dI}{dt}$ here emf is zero when $\frac{dI}{dt} = 0$; $\frac{dI}{dt} = 2t e^{-t} - t^2 e^{-t} = 0$; i.e., $t e^{-t}(t - 2) = 0 \Rightarrow t \neq \infty$ and $t \neq 0$; $\therefore t = 2$ sec.

   Step Solution:

    1.  Identify Condition for Zero EMF: The induced EMF in an inductor is $\varepsilon = -L \frac{dI}{dt}$. For $\varepsilon$ to be zero, the rate of change of current $\frac{dI}{dt}$ must be zero.

    2.  Differentiate the Current Function: Use the Product Rule $(\frac{d}{dt}[uv] = u'v + uv')$ on $I = t^2 e^{-t}$.

    3.  Perform Calculation: $\frac{dI}{dt} = (2t)(e^{-t}) + (t^2)(-e^{-t}) = 2te^{-t} - t^2e^{-t}$.

    4.  Set Derivative to Zero: $e^{-t}(2t - t^2) = 0$. Since $e^{-t}$ is never zero, we solve $2t - t^2 = 0$.

    5.  Solve for t: $t(2 - t) = 0$. This gives $t = 0$ or $t = 2$ seconds. Since $t=0$ is the starting point, the time at which it becomes zero again is $t = 2$ sec.

   Difficulty Level: Medium

   Concept Name: Faraday’s Law of Induction and Product Rule of Differentiation.

   Short cut solution: EMF is zero at the maxima/minima of the current. For $t^n e^{-at}$, the maximum occurs at $t = n/a$. Here $n=2$ and $a=1$, so $t = 2/1 = 2$ sec.

Question 96

   Question: A metal ring is held horizontally and bar magnet is dropped through the ring with its length along the axis of the ring. The acceleration of the falling magnet is.

   Options: 

       A. more than g

       B. equal to g

       C. less than g

       D. either(a) or (c)

   Correct Answer: C

   Year: 1996

   Solution (as Given in the Source): When the magnet is dropped through the ring, an induced current is developed into the ring in the direction opposing the motion of magnet (Lenz's law). Therefore this induced current decreases the acceleration of bar magnet.

   Step Solution:

    1.  Analyze Magnetic Flux: As the magnet falls toward the ring, the magnetic flux passing through the ring increases.

    2.  Apply Lenz's Law: An induced current is generated in the ring to create a magnetic field that opposes the change in flux (the approach of the magnet).

    3.  Identify the Opposing Force: This induced magnetic field exerts an upward repulsive force on the falling magnet.

    4.  Determine Net Force: The net downward force on the magnet is $F_{net} = mg - F_{magnetic}$.

    5.  Compare Acceleration: Since the net force is less than $mg$, the acceleration $a = F_{net}/m$ must be less than $g$.

   Difficulty Level: Easy

   Concept Name: Lenz’s Law.

   Short cut solution: Lenz's law always acts to "oppose the cause." Since the cause is the magnet falling due to gravity, the induced effect will push it up, making $a < g$.

Question 98

   Question: When the key K is pressed at time t = 0, then which of the following statement about the current I in the resistor AB of the given circuit is true?

   Options: 

       A. I oscillates between 1 mA and 2 mA

       B. at $t = 0, I = 2$ mA and with time it goes to 1 mA

       C. $I = 1$ mA at all t

       D. $I = 2$ mA at all t

   Correct Answer: B

   Year: 1995

   Solution (as Given in the Source): Initially, the current will pass through the capacitor (and not through the resistance which is parallel to the capacitor). So effective resistance in the circuit is $R_{AB}$. Therefore the current in the resistor is 2 mA. After some time, the capacitor will become fully charged and will be in its steady state. Now no current will pass through the capacitor and the effective resistance of the circuit will be $(1000 + 1000) = 2000 \Omega$. Therefore current in the resistor $= V/R = 2/2000 = 1 \times 10^{-3}$ A = 1 mA.

   Step Solution:

    1.  Identify Initial State ($t = 0$): At the instant the switch is closed, an uncharged capacitor acts as a short circuit (zero resistance).

    2.  Calculate Initial Current: The current bypasses the resistor parallel to the capacitor. With $V = 2V$ and $R_{AB} = 1000 \Omega$, $I_0 = \frac{2V}{1000\Omega} = 2$ mA.

    3.  Identify Steady State ($t \rightarrow \infty$): Once fully charged, the capacitor acts as an open circuit (infinite resistance), blocking current flow through its branch.

    4.  Calculate Steady State Resistance: The current must now flow through both resistors in series. $R_{total} = 1000 \Omega + 1000 \Omega = 2000 \Omega$.

    5.  Calculate Final Current: $I_{\infty} = \frac{2V}{2000\Omega} = 1 \times 10^{-3}$ A = 1 mA.

   Difficulty Level: Medium

   Concept Name: Transient Response of RC Circuits.

   Short cut solution: Capacitor is a "wire" at $t=0$ and a "break" at $t=\infty$. Initial $R = 1k\Omega \rightarrow I = 2$ mA. Final $R = 2k\Omega \rightarrow I = 1$ mA.

Question 99

   Question: The current I in an A.C. circuit with inductance coil varies with time according to the graph given below. Which one of the following graphs gives the variation of voltage with time?

   Options: 

       A.

       B. 

       C.

       D.

   Correct Answer: A

   Year: 1994

   Solution (as Given in the Source): In an A.C. circuit with inductance coil, the voltage V leads the current I by a phase difference of $90^\circ$. Or the current I lags behind the voltage V by a phase difference of $90^\circ$. Thus the voltage goes on decreasing with the increase in time as shown in the graph (a).

   Step Solution:

    1.  Identify the Phase Relationship: In a pure inductive circuit, voltage ($V$) leads current ($I$) by a phase angle of $90^\circ$ ($\pi/2$ radians).

    2.  Analyze Current Graph: The current starts at zero ($I = I_0 \sin \omega t$), which means it is a sine function.

    3.  Determine Voltage Function: Since voltage leads by $90^\circ$, its function is $V = V_0 \sin(\omega t + 90^\circ) = V_0 \cos \omega t$.

    4.  Evaluate at $t=0$: At the starting point, $V = V_0 \cos(0) = V_0$, meaning the voltage is at its maximum positive value.

    5.  Observe Trend: As time increases from zero, the cosine function immediately begins to decrease, which matches the behavior in graph A.

   Difficulty Level: Medium

   Concept Name: Phase Relationship in an Inductive AC Circuit.

   Short cut solution: In a pure inductor, voltage leads current by $90^\circ$; if the current starts at zero and goes up, the voltage must start at its positive peak and go down.

Question 100

   Question: A straight line conductor of length 0.4 m is moved with a speed of 7 m/s perpendicular to a magnetic field of intensity 0.9 Wb/m². The induced e.m.f. across the conductor is

   Options: 

       A. 5.04 V

       B. 25.2 V

       C. 1.26 V

       D. 2.52 V

   Correct Answer: D

   Year: 1995

   Solution (as Given in the Source): Length of conductor $(l) = 0.4 \text{ m}$; Speed $= 7 \text{ m/s}$ and magnetic field $(B) = 0.9 \text{ Wb/m}^2$. Induced e.m.f. $(\varepsilon) = Blv = 0.9 \times 0.4 \times 7 = 2.52 \text{ V}$.

   Step Solution:

    1.  Identify Given Data: Length $l = 0.4 \text{ m}$, velocity $v = 7 \text{ m/s}$, and magnetic field $B = 0.9 \text{ T}$.

    2.  Check Orientation: The conductor moves perpendicular to the field, meaning the angle $\theta = 90^\circ$.

    3.  Apply Motional EMF Formula: Use $\varepsilon = Blv \sin \theta$.

    4.  Simplify Calculation: Since $\sin 90^\circ = 1$, the formula reduces to $\varepsilon = Blv$.

    5.  Final Multiplication: $\varepsilon = 0.9 \times 0.4 \times 7 = 0.36 \times 7 = 2.52 \text{ V}$.

   Difficulty Level: Easy

   Concept Name: Motional EMF.

   Short cut solution: Directly multiply the provided values for length, speed, and field: $0.4 \times 7 \times 0.9 = 2.52 \text{ V}$.

Question 107

   Question: The total charge, induced in a conducting loop when it is moved in magnetic field depends on

   Options: 

       A. the rate of change of magnetic flux

       B. initial magnetic flux only

       C. the total change in magnetic flux

       D. final magnetic flux only

   Correct Answer: C

   Year: 1992

   Solution (as Given in the Source): $q = \int i dt = \frac{1}{R} \int \varepsilon dt = (\frac{-d\phi}{dt}) \frac{1}{R} \int dt = \frac{1}{R} \int d\phi$. Hence total charge induced in the conducting loop depend upon the total change in magnetic flux. As the emf or iR depends on rate of change of $\phi$, charge induced depends on change of flux.

   Step Solution:

    1.  Relate Charge to Current: Total charge $q$ is the integral of current over time, $q = \int I dt$.

    2.  Substitute Ohm's Law: Since $I = \varepsilon / R$, then $q = \frac{1}{R} \int \varepsilon dt$.

    3.  Apply Faraday's Law: Substitute $\varepsilon = d\phi/dt$ into the integral: $q = \frac{1}{R} \int (d\phi/dt) dt$.

    4.  Simplify the Integral: The $dt$ terms cancel, leaving $q = \frac{1}{R} \int d\phi$.

    5.  Conclude Dependence: The integral $\int d\phi$ represents the total change in flux ($\Delta\phi$), proving that $q = \Delta\phi/R$, which is independent of time or rate.

   Difficulty Level: Medium

   Concept Name: Induced Charge in Electromagnetic Induction.

   Short cut solution: Induced EMF depends on the rate of flux change, but induced charge depends only on the total change in flux divided by resistance ($q = \Delta\phi/R$).

Question 108

   Question: A rectangular coil of 20 turns and area of cross-section 25 sq. cm has a resistance of 100Ω. If a magnetic field which is perpendicular to the plane of coil changes at a rate of 1000 tesla per second, the current in the coil is.

   Options: 

       A. 1 A

       B. 50 A

       C. 0.5 A

       D. 5 A

   Correct Answer: C

   Year: 1992

   Solution (as Given in the Source): 

    $i = \frac{\varepsilon}{R} = \frac{\frac{NA dB}{dt}}{R} = \frac{20 \times (25 \times 10^{-4}) \times 1000}{100} = 0.5 \text{ A}$

   Step Solution:

    1.  Identify parameters: $N = 20$, $A = 25 \text{ cm}^2 = 25 \times 10^{-4} \text{ m}^2$, $R = 100 \Omega$, and $\frac{dB}{dt} = 1000 \text{ T/s}$.

    2.  Apply Faraday's Law for EMF: $\varepsilon = N \frac{d\Phi}{dt}$. Since $A$ is constant and perpendicular to $B$, $\varepsilon = NA \frac{dB}{dt}$.

    3.  Calculate EMF: $\varepsilon = 20 \times (25 \times 10^{-4}) \times 1000 = 50 \text{ V}$.

    4.  Relate EMF to Current: Use Ohm’s Law, $i = \frac{\varepsilon}{R}$.

    5.  Final Calculation: $i = \frac{50}{100} = 0.5 \text{ A}$.

   Difficulty Level: Medium

   Concept Name: Faraday’s Law of Induction and Ohm’s Law.

   Short cut solution: Directly use $i = \frac{NA(\frac{dB}{dt})}{R}$. Multiplying $20 \times 0.0025 \times 1000 = 50$. Then $50/100 = 0.5 \text{ A}$.

Question 109

   Question: Faraday's laws are consequence of conservation of.

   Options: 

       A. energy

       B. energy and magnetic field

       C. charge

       D. magnetic field

   Correct Answer: A

   Year: 1991

   Solution (as Given in the Source): According to Faraday's laws, it is the conservation of energy.

   Step Solution:

    1.  State Faraday's Law: An induced EMF is created by a change in magnetic flux.

    2.  Recall Lenz's Law: The direction of induced current opposes the change in flux that produced it.

    3.  Analyze Work Done: To maintain the change in flux, mechanical work must be done against this opposition.

    4.  Identify Energy Transformation: This mechanical work is converted into the electrical energy of the induced current.

    5.  Conclude: If the induced current did not oppose the change, energy would be created from nothing, violating the Law of Conservation of Energy.

   Difficulty Level: Easy

   Concept Name: Conservation of Energy (linked via Lenz's Law).

   Short cut solution: Lenz's Law, which determines the direction in Faraday's Law, is fundamentally required to satisfy the Law of Conservation of Energy.

Question 112

   Question: A magnetic field of $2 \times 10^{-2} \text{ T}$ acts at right angles to a coil of area $100 \text{ cm}^2$, with 50 turns. The average e.m.f. induced in the coil is 0.1 V, when it is removed from the field in t sec. The value of t is.

   Options: 

       A. 10 s

       B. 0.1 s

       C. 0.01 s

       D. 1 s

   Correct Answer: B

   Year: 1991

   Solution (as Given in the Source): 

    $\varepsilon = \frac{- (\phi_2 - \phi_1)}{t} = \frac{- (0 - NBA)}{t} = \frac{NBA}{t}$

    $t = \frac{NBA}{\varepsilon} = \frac{50 \times 2 \times 10^{-2} \times 10^{-2}}{0.1} = 0.1 \text{ s}$

   Step Solution:

    1.  List given values: $B = 2 \times 10^{-2} \text{ T}$, $A = 100 \text{ cm}^2 = 10^{-2} \text{ m}^2$, $N = 50$, and average $\varepsilon = 0.1 \text{ V}$.

    2.  Determine Change in Flux: Initial flux $\phi_1 = NBA$; Final flux $\phi_2 = 0$ (since it is removed from the field).

    3.  Apply Average EMF formula: $|\varepsilon| = \frac{\Delta\phi}{\Delta t} = \frac{NBA}{t}$.

    4.  Rearrange for time (t): $t = \frac{NBA}{\varepsilon}$.

    5.  Substitute and calculate: $t = \frac{50 \times (2 \times 10^{-2}) \times 10^{-2}}{0.1} = \frac{0.01}{0.1} = 0.1 \text{ s}$.

   Difficulty Level: Medium

   Concept Name: Average Induced EMF and Magnetic Flux.

   Short cut solution: Use $t = \frac{NBA}{\varepsilon}$. Calculate numerator: $50 \times 0.02 \times 0.01 = 0.01$. Then $0.01 / 0.1 = 0.1 \text{ s}$.