Electromagnetic Induction
The induced emf can be produced in a coil by
A. moving the coil with uniform speed inside uniform magnetic field
B. moving the coil with non uniform speed inside uniform magnetic field
C. rotating the coil inside the uniform magnetic field
D. changing the area of the coil inside the uniform magnetic field
Choose the correct answer from the options given below:
A coil is placed in magnetic field such that plane of coil is perpendicular to the direction of magnetic field. The magnetic flux through a coil can be changed :
A. By changing the magnitude of the magnetic field within the coil.
B. By changing the area of coil within the magnetic field.
C. By changing the angle between the direction of magnetic field and the plane of the coil.
D. By reversing the magnetic field direction abruptly without changing its magnitude.
Choose the most appropriate answer from the options given below :
Spherical insulating ball and a spherical metallic ball of same size and mass are dropped from the same height. Choose the correct statement out of the following
{Assume negligible air friction}
A square loop of area 25 cm$^2$ has a resistance of 10 $\Omega$. The loop is placed in uniform magnetic field of magnitude 40.0 T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0 sec, will be
Find the mutual inductance in the arrangement, when a small circular loop of wire of radius '$R$' is placed inside a large square loop of wire of side $L$ $(L \gg R)$. The loops are coplanar and their centres coincide :
A wire of length 1m moving with velocity 8 m/s at right angles to a magnetic field of 2T. The magnitude of induced emf, between the ends of wire will be __________.
A metallic rod of length 'L' is rotated with an angular speed of '$\omega$' normal to a uniform magnetic field 'B' about an axis passing through one end of rod as shown in figure. The induced emf will be :
A conducting circular loop of radius $\frac{10}{\sqrt\pi}$ cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.25 s is :
Take $\pi=\frac{22}{7}$
Explanation:
$\omega = 210 \cdot \frac{2\pi \mathrm{rad}}{60 \mathrm{s}} = 22 \mathrm{rad/s}$
Now, we can find the linear velocity $v$ of the tip of the rod:
$v = \omega r$
where $r$ is the length of the rod (0.2 m).
$v = 22 \mathrm{rad/s} \cdot 0.2 \mathrm{m} = 4.4 \mathrm{m/s}$
Now, we can find the emf developed between the center and the ring using the formula:
$\epsilon = \frac{1}{2} B\ell v$
where $B$ is the magnetic field (0.2 T), $\ell$ is the length of the rod (0.2 m), and $v$ is the linear velocity (4.4 m/s).
$\epsilon = \frac{1}{2} \cdot 0.2 \mathrm{T} \cdot 0.2 \mathrm{m} \cdot 4.4 \mathrm{m/s} = 0.088 \mathrm{V}$
To express this value in mV, we can simply multiply it by 1000:
$\epsilon = 0.088 \mathrm{V} \cdot 1000 = 88 \mathrm{mV}$
So the emf developed between the center and the ring is 88 mV.
An insulated copper wire of 100 turns is wrapped around a wooden cylindrical core of the cross-sectional area $24 \mathrm{~cm}^{2}$. The two ends of the wire are connected to a resistor. The total resistance in the circuit is $12 ~\Omega$. If an externally applied uniform magnetic field in the core along its axis changes from $1.5 \mathrm{~T}$ in one direction to $1.5 ~\mathrm{T}$ in the opposite direction, the charge flowing through a point in the circuit during the change of magnetic field will be ___________ $\mathrm{mC}$.
Explanation:
The magnetic flux through the circuit is proportional to the magnetic field through the core, so we can write $\phi=NBA$, where $N$ is the number of turns in the loop, $B$ is the magnetic field through the core, and $A$ is the cross-sectional area of the core.
As the magnetic field changes from $1.5\mathrm{~T}$ in one direction to $-1.5\mathrm{~T}$ in the opposite direction, the change in magnetic flux is $\Delta\phi=2NBA$.
The induced emf drives a current $I$ through the resistor in the circuit, and the current and the resistance are related by Ohm's law, which is $I=\mathcal{E}/R$. Substituting the expression for $\mathcal{E}$ into this equation, we get $I=-d\phi/dtR$.
The charge $Q$ that flows through the circuit during the change in magnetic field is given by $Q=\int Idt$. Substituting the expression for $I$ into this equation and integrating with respect to time, we get $Q=-\Delta\phi/R$, where $\Delta\phi$ is the change in magnetic flux and $R$ is the resistance of the circuit.
Substituting the given values into this expression, we get:
$Q=-\frac{2NBA}{R}=-\frac{2(100)(1.5)(24\times10^{-4})}{12}=-0.06\mathrm{~C}=-60\mathrm{~mC}$
Therefore, the charge flowing through a point in the circuit during the change of magnetic field is $60\mathrm{~mC}$, which is the same as the provided answer.
A conducting circular loop is placed in a uniform magnetic field of $0.4 \mathrm{~T}$ with its plane perpendicular to the field. Somehow, the radius of the loop starts expanding at a constant rate of $1 \mathrm{~mm} / \mathrm{s}$. The magnitude of induced emf in the loop at an instant when the radius of the loop is $2 \mathrm{~cm}$ will be ___________ $\mu \mathrm{V}$.
Explanation:
The problem involves a conducting circular loop placed in a uniform magnetic field with its plane perpendicular to the field. The radius of the loop is expanding at a constant rate, and we are asked to find the magnitude of the induced emf in the loop at an instant when the radius of the loop is $2 \mathrm{~cm}$.
The magnetic flux through a circular loop of radius $r$ and area $A = \pi r^2$ placed in a uniform magnetic field $B$ perpendicular to the plane of the loop is given by:
$\Phi_B = B A = B \pi r^2$
The induced emf in the loop is given by Faraday's law of electromagnetic induction:
$\mathcal{E} = -\frac{d\Phi_B}{dt}$
In this case, the radius of the loop is expanding at a constant rate of $10^{-3} \mathrm{~m/s}$, which means that the rate of change of the area of the loop is:
$\frac{dA}{dt} = \frac{d}{dt} (\pi r^2) = 2 \pi r \frac{dr}{dt} = 2 \pi (0.02 \mathrm{~m}) (10^{-3} \mathrm{~m/s}) = 4 \times 10^{-5} \mathrm{~m^2/s}$
The magnetic flux through the loop is changing at this rate, and the induced emf in the loop is given by:
$\mathcal{E} = \left|\frac{d\Phi_B}{dt}\right| = \left|\frac{dB}{dt} \frac{dA}{dt}\right| = \left|B \frac{dA}{dt}\right| = \left|0.4 \mathrm{~T} \times 4 \times 10^{-5} \mathrm{~m^2/s}\right| = 16 \pi \mu \mathrm{V}$
Therefore, the magnitude of the induced emf in the loop at an instant when the radius of the loop is $2 \mathrm{~cm}$ is $50.24 $ $ \simeq $ 50 $\mu \mathrm{V}$.
A metallic cube of side $15 \mathrm{~cm}$ moving along $y$-axis at a uniform velocity of $2 \mathrm{~ms}^{-1}$. In a region of uniform magnetic field of magnitude $0.5 \mathrm{~T}$ directed along $z$-axis. In equilibrium the potential difference between the faces of higher and lower potential developed because of the motion through the field will be _________ mV.
Explanation:
$ \begin{aligned} \Delta V & =(E . d) \\\\ & =(V B) \times 0.15 \\\\ & =2 \times \frac{1}{2} \times 0.15 \mathrm{~V} \\\\ & =0.15 \mathrm{~V}=150 \mathrm{mV} \end{aligned} $
The magnetic field B crossing normally a square metallic plate of area $4 \mathrm{~m}^{2}$ is changing with time as shown in figure. The magnitude of induced emf in the plate during $\mathrm{t}=2 s$ to $\mathrm{t}=4 s$, is __________ $\mathrm{mV}$.
Explanation:
$ \begin{aligned} & \varepsilon=\left|\frac{\mathrm{d} \phi}{\mathrm{dt}}\right|=\frac{\mathrm{d}(\mathrm{BA})}{\mathrm{dt}}=\frac{\mathrm{AdB}}{\mathrm{dt}} \\\\ & \varepsilon=\frac{4 \mathrm{~d}(2 \mathrm{t})}{\mathrm{dt}}=4 \times 2=8 \mathrm{mV} \end{aligned} $
A square loop of side $2.0 \mathrm{~cm}$ is placed inside a long solenoid that has 50 turns per centimetre and carries a sinusoidally varying current of amplitude $2.5 \mathrm{~A}$ and angular frequency $700 ~\mathrm{rad} ~\mathrm{s}^{-1}$. The central axes of the loop and solenoid coincide. The amplitude of the emf induced in the loop is $x \times 10^{-4} \mathrm{~V}$. The value of $x$ is __________.
$ \text { (Take, } \pi=\frac{22}{7} \text { ) } $
Explanation:
In this problem, a square loop is inside a long solenoid, and there's a varying current flowing through the solenoid. Because the current is changing, it induces a changing magnetic field inside the solenoid.
According to Faraday's law of electromagnetic induction, a changing magnetic field will induce an electromotive force (emf) in a loop placed in that field. In this case, the loop is the square loop inside the solenoid.
The formula used here is based on Faraday's law, which states that the induced emf in a loop is equal to the rate of change of magnetic flux through the loop. This is given by:
$ \text{emf} = -\frac{d \Phi}{dt} $
where $\Phi$ is the magnetic flux.
The magnetic field inside a solenoid is given by $B = \mu_0 n I$, where $\mu_0$ is the permeability of free space, $n$ is the number of turns per unit length in the solenoid, and $I$ is the current through the solenoid.
The magnetic flux through the square loop is then given by $\Phi = B \cdot A = \mu_0 n I A$, where $A$ is the area of the loop.
When the current is sinusoidal, i.e., $I(t) = I_0 \sin(\omega t)$, its derivative with respect to time is $dI/dt = I_0 \omega \cos(\omega t)$, where $\omega$ is the angular frequency.
Hence, the rate of change of flux becomes:
$ \frac{d \Phi}{dt} = \mu_0 n A \frac{dI}{dt} = \mu_0 n A I_0 \omega \cos(\omega t) $
The emf, which is equal to the negative of the rate of change of flux, will have a maximum value (the amplitude) when $\cos(\omega t) = 1$, giving:
$
\text{Emf amplitude} = \mu_0 n A I_0 \omega$
$ = 4\pi \times 10^{-7} \, \text{T m/A} \times \left(\frac{50}{10^{-2}}\right) \, \text{turns/m} \times (2 \times 10^{-2} \, \text{m})^2 \times 2.5 \, \text{A} \times 700 \, \text{rad/s}
$
which simplifies to:
$ \text{Emf amplitude} = 44 \times 10^{-4} \, \text{V} $
So, the value of $x$ in the question is $44$
A 1 m long metal rod XY completes the circuit as shown in figure. The plane of the circuit is perpendicular to the magnetic field of flux density 0.15 T. If the resistance of the circuit is 5$\Omega$, the force needed to move the rod in direction, as indicated, with a constant speed of 4 m/s will be ____________ 10$^{-3}$ N.
Explanation:
To move the rod with a constant velocity $v=4 \mathrm{~m} \mathrm{~s}^{-1}$
$\mathrm{F}_{\text {net }}$ on the rod should be zero.
$ \begin{aligned} F & =B i l=0.15\left(\frac{B l v}{R}\right) l \\\\ & =0.15\left(\frac{0.15 \times 1 \times 4}{5}\right) \times 1 \\\\ & =0.03 \times 0.15 \times 4 \\\\ & =180 \times 10^{-4} \mathrm{~N} \\\\ & =18 \times 10^{-3} \mathrm{~N} \end{aligned} $
Two concentric circular coils with radii $1 \mathrm{~cm}$ and $1000 \mathrm{~cm}$, and number of turns 10 and 200 respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be ___________ $\times 10^{-8} \mathrm{H}$. (Take, $\pi^{2}=10$ )
Explanation:
The magnetic field $B_2$ due to the current $I_2$ in the larger coil with 200 turns is given by:
$B_2 = \frac{N_2 \mu_0 I_2}{2r_2} = \frac{200 \mu_0 I_2}{2 \times 10}$
The magnetic flux $\phi_{1,2}$ through the smaller coil due to this magnetic field is given by:
$\phi_{1,2} = N_1 \vec{B}_2 \cdot \vec{A}_1 = N_1 N_2 \frac{\mu_0 I_2}{2 r_2} \cdot \pi r_1^2$
Since $\phi_{1,2} = MI_2$, we can solve for the mutual inductance $M$:
$M = \frac{N_1 N_2 \frac{\mu_0 I_2}{2 r_2} \cdot \pi r_1^2}{I_2}$
Substituting the given values for $r_1$, $N_1$, $r_2$, and $N_2$:
$M = \frac{10 \times 200 \times 4 \pi \times 10^{-7} \times \pi \times (0.01)^2}{2 \times 10}$
Simplifying the expression, we get:
$M = 4 \times 10^{-8} \mathrm{H}$
So, the mutual inductance between the two concentric coils is $4 \times 10^{-8} \mathrm{H}$.
As per the given figure, if $\frac{\mathrm{dI}}{\mathrm{dt}}=-1 \mathrm{~A} / s$ then the value of $\mathrm{V}_{\mathrm{AB}}$ at this instant will be ____________ $\mathrm{V}$.
Explanation:
From the circuit :
${V_A} - iR - {{Ldi} \over {dt}} - 12 = {V_B}$
$ \Rightarrow {V_A} - {V_B} = 2 \times 12 + 6( - 1) + 12$ volts
$ = 30$ volts
A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field B = 0.8 T. When released the radius of the loop starts shrinking at a constant rate of 2 cms$^{-1}$. The induced emf in the loop at an instant when the radius of the loop is 10 cm will be __________ mV.
Explanation:
$=2 \mathrm{~B} \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}=2 \times \pi \times 0.1 \times 0.8 \times 2 \times 10^{-2}$
$=2 \pi \times 1.6=\mathbf{1 0 . 0 6}~[$ rounding off $\mathbf{1 0 . 0 6}=\mathbf{1 0}]$
Three identical resistors with resistance R = 12$\Omega$ and two identical inductors with self inductance L = 5 mH are connected to an ideal battery with emf of 12 V as shown in figure. The current through the battery long after the switch has been closed will be _____________ A.
Explanation:
After long time, inductors are shorted.
Effective circuit becomes
Current through battery $=\frac{V}{\mathrm{R}_{\mathrm{eq}}}=\frac{12 \mathrm{~V}}{4 \Omega}=3 \mathrm{~A}$
where $\mathrm{R}_{\mathrm{eq}}=3$ resistors in parallel.
The electric current in a circular coil of 2 turns produces a magnetic induction B1 at its centre. The coil is unwound and in rewound into a circular coil of 5 tuns and the same current produces a magnetic induction B2 at its centre. The ratio of ${{{B_2}} \over {{B_1}}}$ is
A small square loop of wire of side $l$ is placed inside a large square loop of wire $\mathrm{L}(\mathrm{L}>>l)$. Both loops are coplanar and their centres coincide at point $\mathrm{O}$ as shown in figure. The mutual inductance of the system is :
A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be :
(Assume the coil to be short circuited.)
Two coils of self inductance L1 and L2 are connected in series combination having mutual inductance of the coils as M. The equivalent self inductance of the combination will be :
A metallic conductor of length 1 m rotates in a vertical plane parallel to east-west direction about one of its end with angular velocity 5 rad s$-$1. If the horizontal component of earth's magnetic field is 0.2 $\times$ 10$-$4 T, then emf induced between the two ends of the conductor is :
The magnetic flux through a coil perpendicular to its plane is varying according to the relation $\phi = (5{t^3} + 4{t^2} + 2t - 5)$ Weber. If the resistance of the coil is 5 ohm, then the induced current through the coil at t = 2 s will be,
For the given circuit the current through battery of 6 V just after closing the switch 'S' will be _________ A.
Explanation:
Just after closing the switch, $i = {6 \over {4 + 2}} = 1\,A$
A conducting circular loop is placed in $X-Y$ plane in presence of magnetic field $\overrightarrow{\mathrm{B}}=\left(3 \mathrm{t}^{3} \,\hat{j}+3 \mathrm{t}^{2}\, \hat{k}\right)$ in SI unit. If the radius of the loop is $1 \mathrm{~m}$, the induced emf in the loop, at time, $\mathrm{t}=2 \mathrm{~s}$ is $\mathrm{n} \pi \,\mathrm{V}$. The value of $\mathrm{n}$ is ___________.
Explanation:
${B_ \bot } = 3{t^2}$
${{d{B_ \bot }} \over {dt}} = 6t - 12$ at $t = 2$
${{d{\phi _1}} \over {dt}} = 12 \times \pi {(1)^2} = 12\pi $
In a coil of resistance $8 \,\Omega$, the magnetic flux due to an external magnetic field varies with time as $\phi=\frac{2}{3}\left(9-t^{2}\right)$. The value of total heat produced in the coil, till the flux becomes zero, will be _____________ $J$.
Explanation:
$R = 8\,\Omega $
$\phi = {2 \over 3}(9 - {t^2})$
At $t = 3$, $\phi = 0$
$\varepsilon = \left| { - {{d\phi } \over {dt}}} \right| = {4 \over 3}t$
$H = \int_0^3 {{{{V^2}} \over R}dt = \int_0^3 {{1 \over 8} \times {{16} \over 9}{t^2}dt} } $
$ = {2 \over 9} \times \left( {{{{t^3}} \over 3}} \right)_0^3 = {2 \over {9 \times 3}} \times 27 = 2\,J$
Magnetic flux (in weber) in a closed circuit of resistance 20 $\Omega$ varies with time t(s) at $\phi$ = 8t2 $-$ 9t + 5. The magnitude of the induced current at t = 0.25 s will be ____________ mA.
Explanation:
$R = 20\,\Omega $
$\phi = 8{t^2} - 9t + 5$
$\varepsilon = \left| { - {{d\phi } \over {dt}}} \right| = |16t - 9| = |16(0.25) - 9| = 5$
$i = {\varepsilon \over R} = {5 \over {20}} = 0.25\,A = {{0.25} \over {{{10}^3}}} \times {10^3}\,A = 250\,mA$
A metallic rod of length 20 cm is placed in North-South direction and is moved at a constant speed of 20 m/s towards East. The horizontal component of the Earth's magnetic field at that place is 4 $\times$ 10$-$3 T and the angle of dip is 45$^\circ$. The emf induced in the rod is ___________ mV.
Explanation:
$E = Blv$
$ = 4 \times {10^{ - 3}} \times {{20} \over {100}} \times 20$ Volts
$ = 16$ mV
A 10 $\Omega$, 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 $\mu$s. The average e.m.f. induced in the coil is ____________ V.
Explanation:
Initially current, ${I_0} = {{20} \over {10}} = 2A$ (when initially switch closed)
average emf induced in coil $ = {{Ldi} \over {dt}}$
$ \Rightarrow {e_{avg}} = {{20 \times {{10}^{ - 3}} \times (2 - 0)} \over {100 \times {{10}^{ - 6}}}}$
${e_{avg}} = 400\,V$
The current in a coil of self inductance 2.0 H is increasing according to I = 2 sin(t2) A. The amount of energy spent during the period when current changes from 0 to 2 A is ____________ J.
Explanation:
$U = {1 \over 2}L{I^2}$
$ = {1 \over 2}2 \times {2^2} = 4$ J
A circular coil of 1000 turns each with area 1m2 is rotated about its vertical diameter at the rate of one revolution per second in a uniform horizontal magnetic field of 0.07T. The maximum voltage generation will be ___________ V.
Explanation:
${V_{\max }} = NAB\omega $
$ = 1000 \times 1 \times 0.07 \times (2\pi \times 1)$
$ \simeq 440$ volts
A current is induced in the coil because $\overrightarrow B $ is :
Two resistors R1 and R2 are connected across the ends of the rails. There is a uniform magnetic field $\overrightarrow B $ pointing into the page. An external agent pulls the bar to the left at a constant speed v.
The correct statement about the directions of induced currents I1 and I2 flowing through R1 and R2 respectively is :
[mp = 1.67 $\times$ 10$-$27 kg, e = 1.6 $\times$ 10$-$19C, Speed of light = 3 $\times$ 108 m/s]
Explanation:
Number of revolution = n
$n[2 \times {q_P} \times V] = {1 \over 2}{m_P} \times v_P^2$
$n[2 \times 1.6 \times {10^{ - 19}} \times 12 \times {10^3}]$
$ = {1 \over 2} \times 1.67 \times {10^{ - 27}} \times {\left[ {{{3 \times {{10}^8}} \over 6}} \right]^2}$
n(38.4 $\times$ 10$-$16) = 0.2087 $\times$ 10$-$11
n = 543.4
Explanation:
N = 20, $\omega$ = 50, B = 3 $\times$ 10$-$2 T
$\varepsilon $ = 20 $\times$ 50 $\times$ $\pi$ $\times$ (0.08)2 $\times$ 3 $\times$ 10$-$2 = 60.28 $\times$ 10$-$2
Rounded off to nearest integer = 60
The magnitude of current through R = 2$\Omega$ resistor at t = 5 s is ___________ mA.
Explanation:
$\left| i \right| = {{\left| \in \right|} \over R}$ = 10t + 10 mA
at t = 5
$\left| i \right|$ = 60 mA
$B = {4 \over \pi } \times {10^{ - 3}}T\left( {1 - {t \over {100}}} \right)$
The energy dissipated by the coil before the magnetic field is switched off completely is E = ___________ mJ.
Explanation:
$\phi = {4 \over \pi } \times {10^{ - 3}}\left( {1 - {t \over {100}}} \right).\pi {R^2}$
$\phi = 4 \times {10^{ - 3}} \times {(1)^2}\left( {1 - {t \over {100}}} \right)$
$\varepsilon = {{ - d\phi } \over {dt}}$
$\varepsilon = {{ - d} \over {dt}}\left( {4 \times {{10}^{ - 3}}\left( {1 - {t \over {100}}} \right)} \right)$
$\varepsilon = 4 \times {10^{ - 3}}\left( {{1 \over {100}}} \right) = 4 \times {10^{ - 5}}V$
When B = 0
$1 - {t \over {100}} = 0$
t = 100 sec
Heat $ = {{{\varepsilon ^2}} \over R}t$
Heat $ = {{{{(4 \times {{10}^{ - 5}})}^2}} \over {2 \times {{10}^{ - 6}}}} \times 100$ J
Heat $ = {{16 \times {{10}^{ - 10}} \times 100} \over {2 \times {{10}^{ - 6}}}}$ J
Heat = 0.08 J
Heat = 80 mJ
Explanation:
$ = 3t$
$L\int {di = 3\int {tdt} } $
$Li = {{3{t^2}} \over 2}$
$i = {{3{t^2}} \over {2L}}$
energy, $E = {1 \over 2}L{i^2}$
$ = {1 \over 2}L{\left( {{{3{t^2}} \over {2L}}} \right)^2}$
$ = {1 \over 2} \times {{9{t^4}} \over {4L}}$
$ = {9 \over 8} \times {{{{(4)}^4}} \over {4 \times 2}} = 144$ J