Question 5 (Field at the center)

   Question: A tightly wound 100 turns coil of radius 10cm carries a current of 7A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4 \pi \times 1 0 ^ { - 7 } \mathrm { S I }$ units):

   Options: 

    A. 44 mT

    B. 4.4T

    C. 4.4 mT

    D. 44T

   Correct Answer: C

   Year: NEET 2024

   Solution: The magnitude of magnetic field due to circular coil of N turns is given by $B _ {C} = \frac {\mu_ {0} i N}{2 R} = \frac {4 \pi \times 1 0 ^ {- 7} \times 7 \times 1 0 0}{2 \times 0 . 1} = 4. 4 \times 1 0 ^ {- 3} \mathrm {T} = 4. 4 \mathrm {m T}$.

   Step Solution:

    1.  Identify given values: Number of turns $N = 100$, radius $R = 10 \text{ cm} = 0.1 \text{ m}$, and current $I = 7 \text{ A}$.

    2.  State the formula: Use the formula for the magnetic field at the center of a circular coil: $B = \frac{\mu_0 N I}{2R}$.

    3.  Substitute values: $B = \frac{4\pi \times 10^{-7} \times 100 \times 7}{2 \times 0.1}$.

    4.  Perform calculation: $B = \frac{28\pi \times 10^{-5}}{0.2} = 140\pi \times 10^{-5} \text{ T}$.

    5.  Final result: $B \approx 439.8 \times 10^{-5} \text{ T} \approx 4.4 \times 10^{-3} \text{ T} = 4.4 \text{ mT}$.

   Difficulty Level: Easy

   Concept Name: Magnetic field at the center of a circular coil.

   Short cut solution: Use the approximation $\frac{\mu_0}{2} = 2\pi \times 10^{-7} \approx 6.28 \times 10^{-7}$. Then $B \approx (6.28 \times 10^{-7}) \times (100) \times (7) / (0.1) = 43.96 \times 10^{-4} \approx 4.4 \text{ mT}$.

Question 11 (Field at the center)

   Question: A closely packed coil having 1000 turns has an average radius of 62.8 cm. If current carried by the wire of the coil is 1A, the value of magnetic field produced at the centre of the coil will be (permeability of free space $= 4 \pi \times 1 0 ^ { - 7 } \mathrm { H / m }$) nearly.

   Options: 

    A. $10^{-3}$T

    B. $10^{-1}$T

    C. $10^{-2}$T

    D. $10^2$T

   Correct Answer: A

   Year: NEET Re-2022

   Solution: $B = \frac {\mu_ {0} n l}{2 R} = \frac {4 \pi \times 1 0 ^ {- 7} \times 1 0 0 0 \times 1}{2 \times 6 2 . 8 \times 1 0 ^ {- 2}} = 1 0 ^ {- 3} T$.

   Step Solution:

    1.  Identify given values: $N = 1000$, $R = 62.8 \text{ cm} = 0.628 \text{ m}$, and $I = 1 \text{ A}$.

    2.  State the formula: $B = \frac{\mu_0 N I}{2R}$.

    3.  Substitute values: $B = \frac{4\pi \times 10^{-7} \times 1000 \times 1}{2 \times 0.628}$.

    4.  Simplify numerical values: Since $\pi \approx 3.14$, then $4\pi \approx 12.56$. Note that $2 \times 0.628 = 1.256$.

    5.  Final calculation: $B = \frac{12.56 \times 10^{-4}}{1.256} = 10 \times 10^{-4} = 10^{-3} \text{ T}$.

   Difficulty Level: Easy

   Concept Name: Magnetic field at the center of a circular coil.

   Short cut solution: Recognize that $R = 62.8 \text{ cm} = 20\pi \text{ cm}$, so $2R = 40\pi \text{ cm} = 0.4\pi \text{ m}$. Thus, $B = \frac{4\pi \times 10^{-7} \times 1000}{0.4\pi} = \frac{4}{0.4} \times 10^{-4} = 10^{-3} \text{ T}$.

Question 24 (Field at the center)

   Question: A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is R. The total magnetic field at the centre P at the loop is

   Options: 

    A. Zero

    B. $3 \mu _ { 0 } \ i / 3 2 \mathrm { R }$, outward

    C. $3 \mu _ { 0 } \ i / 3 2 \ R$, inward

    D. $\frac { \mu _ { 0 } { \mathrm { i } } } { 2 \mathrm { R } }$, inward

   Correct Answer: A

   Year: OD NEET 2019

   Solution: Magnetic field due to $i_1 = \frac {\mu_ {0} \dot {\mathbf {i}} _ {1}}{2 R} \left(\frac {\theta_ {1}}{2 \pi}\right)$ (Into the plane) and $i_2 = \frac {\mu_ {0} \dot {\mathbf {i}} _ {2}}{2 R} \left(\frac {\theta_ {2}}{2 \pi}\right)$ (Out of the plane). For parallel combination $\frac{i_1}{i_2} = \frac{l_2}{l_1} = \frac{1}{3}$. Net field $= \frac {\mu_ {0}}{2 R} [ \frac {3 i _ {1}}{4} - \frac {i _ {2}}{4} ] = \frac {\mu_ {0}}{2 R} [ \frac {3 i _ {1}}{4} - \frac {3 i _ {1}}{4} ] = 0$.

   Step Solution:

    1.  Analyze current distribution: The current $i$ splits into two arcs of length $l_1$ and $l_2$ in parallel.

    2.  Relate current and length: For parallel branches, $V_1 = V_2 \implies I_1 R_1 = I_2 R_2$. Since resistance $R \propto l$, we have $I_1 l_1 = I_2 l_2$.

    3.  Relate length and angle: For a circle, $l = R\theta$, so $I_1 (R \theta_1) = I_2 (R \theta_2)$, which simplifies to $I_1 \theta_1 = I_2 \theta_2$.

    4.  Calculate individual fields: The field from each arc at the center is $B = \frac{\mu_0 I \theta}{4 \pi R}$.

    5.  Calculate net field: The fields are in opposite directions (one inward, one outward), so $B_{net} = B_1 - B_2 = \frac{\mu_0}{4 \pi R} (I_1 \theta_1 - I_2 \theta_2)$. Since $I_1 \theta_1 = I_2 \theta_2$, the net field is $0$.

   Difficulty Level: Medium

   Concept Name: Magnetic field due to current-carrying arcs and parallel distribution of current.

   Short cut solution: For any current entering and leaving a uniform circular loop at any two points, the magnetic field at the center is always Zero because the magnetic effect of the shorter arc is perfectly canceled by the larger arc.

Question 30 (Field at the center)

   Question: A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be.

   Options: 

    A. nB

    B. $n^2B$

    C. 2nB

    D. $2n^2B$.

   Correct Answer: B.

   Year: 2016 NEET Phase II.

   Solution: Let $l$ be the length of the wire. Magnetic field at the centre of the loop is

$B = \frac{\mu_0 I}{2R} \therefore B = \frac{\mu_0 \pi I}{l} (\because l = 2 \pi R) \dots (i)$. For $n$ turns, $B' = \frac{\mu_0 n I}{2r} = \frac{\mu_0 n I}{2 (\frac{l}{2n\pi})} = \frac{\mu_0 n^2 \pi I}{l} \dots (ii)$. From eqns (i) and (ii), we get $B' = n^2B$.

   Step Solution:

    1.  Let the total wire length be $l$. For one turn, the radius is $R = \frac{l}{2\pi}$, and the magnetic field is $B = \frac{\mu_0 I}{2R} = \frac{\mu_0 \pi I}{l}$.

    2.  When the same wire is bent into $n$ turns, the new radius $r$ satisfies $l = n(2\pi r)$, which gives $r = \frac{l}{2n\pi}$.

    3.  The magnetic field at the center of a coil with $n$ turns is $B' = \frac{\mu_0 n I}{2r}$.

    4.  Substituting the value of $r$, we get $B' = \frac{\mu_0 n I}{2(\frac{l}{2n\pi})} = \frac{\mu_0 n^2 \pi I}{l}$.

    5.  Comparing this with the initial field, $B' = n^2 \left(\frac{\mu_0 \pi I}{l}\right) = n^2 B$.

   Difficulty Level: Medium

   Concept Name: Magnetic field at the center of a circular coil.

   Short cut solution: For a fixed length of wire bent into $n$ turns, the magnetic field at the center is always proportional to $n^2$, hence $B' = n^2B$.

Question 32 (Field at the center)

   Question: An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has magnitude.

   Options: 

    A. $\frac{\mu_0 n^2 e}{r}$

    B. $\frac{\mu_0 n e}{2r}$

    C. $\frac{\mu_0 n e}{2\pi r}$

    D. Zero.

   Correct Answer: B.

   Year: 2015, Cancelled.

   Solution: Current in the orbit, $I = \frac{e}{T} = \frac{\omega e}{2\pi} = \frac{(2\pi n)e}{2\pi} = ne$. Magnetic field at centre of current carrying circular coil is given by $B = \frac{\mu_0 I}{2r} = \frac{\mu_0 n e}{2r}$.

   Step Solution:

    1.  The moving electron acts as a current loop where current $I$ is the rate of charge flow: $I = \frac{e}{T}$.

    2.  Since the electron makes $n$ rotations per second, the time period $T$ is $\frac{1}{n}$.

    3.  Calculate the equivalent current: $I = \frac{e}{1/n} = ne$.

    4.  Apply the formula for the magnetic field at the center of a circular loop: $B = \frac{\mu_0 I}{2r}$.

    5.  Substitute the current value: $B = \frac{\mu_0 (ne)}{2r} = \frac{\mu_0 ne}{2r}$.

   Difficulty Level: Easy

   Concept Name: Magnetic field due to a rotating charge.

   Short cut solution: Use the relation $I = q \cdot f$, where $f$ is frequency. Here $I = en$. The field at the center is $\frac{\mu_0 I}{2r}$, which immediately gives $\frac{\mu_0 ne}{2r}$.

Question 40 (Field at the center)

   Question: Two similar coils of radius R are lying concentrically with their planes at right angles to each other. The currents flowing in them are I and 2I, respectively. The resultant magnetic field induction at the centre will be.

   Options: 

    A. $\frac{\sqrt{5}\mu_0 I}{2R}$

    B. $\frac{3\mu_0 I}{2R}$

    C. $\frac{\mu_0 I}{2R}$

    D. $\frac{\sqrt{5}\mu_0 I}{R}$.

   Correct Answer: A.

   Year: 2012.

   Solution: Magnetic field induction due to vertical loop at the centre O is $B_1 = \frac{\mu_0 I}{2R}$. It acts in horizontal direction. Magnetic field induction due to horizontal loop at the centre O is $B_2 = \frac{\mu_0 2 I}{2R}$. It acts in vertically upward direction. As $B_1$ and $B_2$ are perpendicular to each other, the resultant is $B_{net} = \sqrt{B_1^2 + B_2^2} = \sqrt{(\frac{\mu_0 I}{2R})^2 + (\frac{\mu_0 2 I}{2R})^2} = \frac{\mu_0 I}{2R} \sqrt{(1)^2 + (2)^2} = \frac{\sqrt{5}\mu_0 I}{2R}$.

   Step Solution:

    1.  Calculate field from the first coil: $B_1 = \frac{\mu_0 I}{2R}$.

    2.  Calculate field from the second coil: $B_2 = \frac{\mu_0 (2I)}{2R}$.

    3.  Identify the direction: Because the planes are perpendicular, the magnetic field vectors $\vec{B_1}$ and $\vec{B_2}$ are also perpendicular to each other.

    4.  Use vector addition for perpendicular vectors: $B_{net} = \sqrt{B_1^2 + B_2^2}$.

    5.  Substitute and solve: $B_{net} = \sqrt{(\frac{\mu_0 I}{2R})^2 + (\frac{2\mu_0 I}{2R})^2} = \frac{\mu_0 I}{2R} \sqrt{1^2 + 2^2} = \frac{\sqrt{5}\mu_0 I}{2R}$.

   Difficulty Level: Medium

   Concept Name: Vector addition of magnetic fields.

   Short cut solution: Since $B \propto I$, if the field due to current $I$ is $B_0$, the field due to $2I$ is $2B_0$. The resultant of two perpendicular fields $B_0$ and $2B_0$ is $\sqrt{B_0^2 + (2B_0)^2} = \sqrt{5}B_0$. Substituting $B_0 = \frac{\mu_0 I}{2R}$ gives $\frac{\sqrt{5}\mu_0 I}{2R}$.

Question 47 (Field at the center)

   Question: Charge $q$ is uniformly spread on a thin ring of radius $R$. The ring rotates about its axis with a uniform frequency $f$ Hz. The magnitude of magnetic induction at the centre of the ring is:

   Options: 

    A. $\frac{\mu_0 q f}{2 \pi R}$

    B. $\frac{\mu_0 q f}{2 R}$

    C. $2f R \mu_0 q$

    D. $\frac{\mu_0 q}{2 \pi \rho R}$

   Correct Answer: B

   Year: 2011 Mains

   Solution: The current flowing in the ring is $I = q f$. The magnetic induction at the centre of the ring is $B = \frac{\mu_0 2 \pi I}{4 \pi R} = \frac{\mu_0 q f}{2 R}$.

   Step Solution:

    1.  Define Current: The moving charge $q$ with frequency $f$ creates an equivalent current $I = \frac{q}{T} = qf$.

    2.  State Formula: The magnetic field at the center of a circular loop is $B = \frac{\mu_0 I}{2R}$.

    3.  Substitute Values: Replace $I$ with $qf$ in the formula: $B = \frac{\mu_0 (qf)}{2R}$.

    4.  Simplify: The expression becomes $B = \frac{\mu_0 q f}{2R}$.

    5.  Final Result: The magnitude of magnetic induction is $\frac{\mu_0 q f}{2 R}$.

   Difficulty Level: Easy

   Concept Name: Magnetic field at the center of a rotating charged ring.

   Short cut solution: Use $I = qf$ directly in the standard loop formula $B = \frac{\mu_0 I}{2R}$ to get $\frac{\mu_0 qf}{2R}$.

Question 49 (Field at the center)

   Question: A thin ring of radius $R$ meter has charge $q$ coulomb uniformly spread on it. The ring rotates about its axis with a constant frequency of $f$ revolutions/s. The value of magnetic induction in $Wb/m^2$ at the centre of the ring is:

   Options: 

    A. $\frac{\mu_0 q f}{2 \pi R}$

    B. $\frac{\mu_0 q}{2 \pi f R}$

    C. $2f R \mu_0 q$

    D. $\frac{\mu_0 q f}{2 R}$

   Correct Answer: D

   Year: 2010

   Solution: Current produced due to circular motion of charge $q$ is $I = qf$. Magnetic field induction at the centre of the ring of radius $R$ is $B = \frac{\mu_0 2 \pi I}{4 \pi R} = \frac{\mu_0 I}{2 R} = \frac{\mu_0 q f}{2 R}$.

   Step Solution:

    1.  Identify equivalents: Revolutions per second is frequency ($f$).

    2.  Calculate current: $I = \text{Charge} \times \text{Frequency} = qf$.

    3.  Apply loop formula: Use $B = \frac{\mu_0 I}{2R}$ for the field at the center.

    4.  Perform substitution: $B = \frac{\mu_0 (qf)}{2R}$.

    5.  Final expression: $B = \frac{\mu_0 q f}{2 R}$.

   Difficulty Level: Easy

   Concept Name: Magnetic field due to a rotating charge.

   Short cut solution: This is identical in logic to Question 47; $I = qf$ leads directly to $\frac{\mu_0 qf}{2R}$.

Question 52 (Field at the center)

   Question: A current loop consists of two identical semicircular parts each of radius $R$, one lying in the $x-y$ plane and the other in $x-z$ plane. If the current in the loop is $i$. The resultant magnetic field due to the two semi circular parts at their common centre is:

   Options: 

    A. $\frac{\mu_0 i}{2 \sqrt{2} R}$

    B. $\frac{\mu_0 i}{2 R}$

    C. $\frac{\mu_0 i}{4 R}$

    D. $\frac{\mu_0 i}{\sqrt{2} R}$

   Correct Answer: A

   Year: 2010 Mains

   Solution: Magnetic field at the centre due to semicircular loop lying in $x-y$ plane, $B_{xy} = \frac{1}{2} \frac{\mu_0 i}{2R}$ in negative $z$ direction. Similarly, field due to loop in $x-z$ plane, $B_{xz} = \frac{1}{2} \frac{\mu_0 i}{2R}$ in negative $y$ direction. Resultant $B = \sqrt{B_{xy}^2 + B_{xz}^2} = \sqrt{(\frac{\mu_0 i}{4R})^2 + (\frac{\mu_0 i}{4R})^2} = \frac{\mu_0 i}{4R} \sqrt{2} = \frac{\mu_0 i}{2 \sqrt{2} R}$.

   Step Solution:

    1.  Calculate field of one arc: A semicircle produces half the field of a full loop: $B_{arc} = \frac{1}{2} (\frac{\mu_0 i}{2R}) = \frac{\mu_0 i}{4R}$.

    2.  Determine directions: The $x-y$ plane arc produces a field along the $z$-axis; the $x-z$ plane arc produces a field along the $y$-axis.

    3.  Identify vector orientation: These two magnetic fields are perpendicular to each other.

    4.  Use vector addition: Resultant $B_{net} = \sqrt{B_1^2 + B_2^2} = \sqrt{(\frac{\mu_0 i}{4R})^2 + (\frac{\mu_0 i}{4R})^2}$.

    5.  Calculate final value: $B_{net} = \frac{\mu_0 i}{4R} \sqrt{2} = \frac{\mu_0 i}{2 \sqrt{2} R}$.

   Difficulty Level: Medium

   Concept Name: Vector addition of magnetic fields from semicircular arcs.

   Short cut solution: Recognize that for two perpendicular fields of equal magnitude $B_0$, the resultant is $\sqrt{2}B_0$. Since $B_0 = \frac{\mu_0 i}{4R}$, the net field is $\frac{\sqrt{2}\mu_0 i}{4R} = \mathbf{\frac{\mu_0 i}{2 \sqrt{2} R}}$.

Question 63 (Field at the center)

   Question: Two circular coils 1 and 2 are made from the same wire but the radius of the 1st coil is twice that of the 2nd coil. What potential difference in volts should be applied across them so that the magnetic field at their centres is the same?

   Options: 

    A. 2

    B. 3

    C. 4

    D. 6

   Correct Answer: C

   Year: 2006

   Solution: The magnetic field at the centre of the coil, $B = \frac{\mu_0 n i}{2 r}$ where $r$ is the radius. $i = \frac{E}{R}$. $R \propto 2 \pi r \implies R = cr$. In the first coil, $B_1 = \frac{\mu_0 n E_1}{2 r_1 (c r_1)} = \frac{\mu_0 n E_1}{2 c r_1^2}$. If $r_1 = 2r_2$, then $B_1 = \frac{\mu_0 n E_1}{2 c (2 r_2)^2} = \frac{\mu_0 n E_1}{8 c r_2^2}$. $B_2 = \frac{\mu_0 n E_2}{2 c r_2^2}$. For $B_1 = B_2$, $\frac{E_1}{4} = E_2 \implies \frac{E_1}{E_2} = 4$.

   Step Solution: 

    1.  Relate resistance to radius: For same wire, resistance $R = \rho \frac{l}{A}$. Length $l = 2\pi r$, so $R \propto r$. Thus, $R_1 = 2R_2$ since $r_1 = 2r_2$.

    2.  State Magnetic Field formula: Use $B = \frac{\mu_0 I}{2r}$ and substitute $I = \frac{V}{R}$ (Ohm's Law).

    3.  Combine formulas: $B = \frac{\mu_0 V}{2rR}$. Since $R \propto r$, the field $B \propto \frac{V}{r^2}$.

    4.  Set fields equal: For $B_1 = B_2$, we must have $\frac{V_1}{r_1^2} = \frac{V_2}{r_2^2}$.

    5.  Calculate ratio: $\frac{V_1}{V_2} = (\frac{r_1}{r_2})^2 = (2)^2 = 4$.

   Difficulty Level: Medium

   Concept Name: Ohm's Law and Magnetic field of a circular coil.

   Short cut solution: Since $B \propto \frac{I}{r}$ and $I \propto \frac{V}{r}$ (because resistance $\propto$ radius), then $B \propto \frac{V}{r^2}$. For $B$ to be constant, $V$ must be proportional to $r^2$. Thus, $(2)^2 = 4$.

Question 71 (Field at the center)

   Question: The magnetic field of given length of wire for single turn coil at its centre is B then its value for two turns coil for the same wire is

   Options: 

    A. B

    B. 2B

    C. 4B

    D. 2B

   Correct Answer: C

   Year: 2002

   Solution: The magnetic field B produced at the centre of a circular coil is given by $B = \frac{\mu_0 N I}{2 r}$. Here $B = \frac{\mu_0 I}{2 r} [N = 1]$.

For same length, $2 \times 2 \pi r' = 2 \pi r \implies r' = \frac{r}{2}$. Magnetic field for two turns ($N = 2$) is $B' = \frac{\mu_0 \times 2 I}{2 r'} = \frac{2 \mu_0 I}{2(r/2)} = \frac{4 \mu_0 I}{2 r} = 4B$.

   Step Solution:

    1.  State initial condition: For $N=1$, $B = \frac{\mu_0 I}{2r}$.

    2.  Determine new radius: Total length $L$ is constant. $L = 2\pi r$ (for $N=1$) and $L = 2(2\pi r')$ (for $N=2$). This gives $r' = \frac{r}{2}$.

    3.  State new field formula: $B' = \frac{\mu_0 N I}{2r'}$ where $N=2$.

    4.  Substitute values: $B' = \frac{\mu_0 (2) I}{2(r/2)}$.

    5.  Calculate final field: $B' = \frac{4 \mu_0 I}{2r} = 4B$.

   Difficulty Level: Easy

   Concept Name: Magnetic field at the center of a circular coil.

   Short cut solution: For a fixed length of wire bent into $N$ turns, the magnetic field at the center $B \propto N^2$. Thus, for $N=2$, $B' = 2^2 B = 4B$.

Question 75 (Field at the center)

   Question: Magnetic field due to 0.1 A current flowing through a circular coil of radius 0.1 m and 1000 turns at the centre of the coil is

   Options: 

    A. $6.28 \times 10^{-4}$ T

    B. $4.31 \times 10^{-2}$ T

    C. $2 \times 10^{-1}$ T

    D. $9.81 \times 10^{-4}$ T

   Correct Answer: A

   Year: 1999

   Solution: $B = \frac{\mu_0 N i}{2 r} = \frac{4 \pi \times 10^{-7} \times 1000 \times 0.1}{2 \times 0.1} = 6.28 \times 10^{-4}$ T.

   Step Solution:

    1.  Identify given values: $I = 0.1$ A, $r = 0.1$ m, and $N = 1000$.

    2.  State the formula: $B = \frac{\mu_0 N I}{2r}$.

    3.  Substitute numerical values: $B = \frac{4\pi \times 10^{-7} \times 1000 \times 0.1}{2 \times 0.1}$.

    4.  Simplify the expression: The $0.1$ in the numerator and denominator cancels out: $B = \frac{4\pi \times 10^{-4}}{2} = 2\pi \times 10^{-4}$.

    5.  Final calculation: $B = 2 \times 3.14 \times 10^{-4} = 6.28 \times 10^{-4}$ T.

   Difficulty Level: Easy

   Concept Name: Magnetic field at the center of a circular coil.

   Short cut solution: Direct substitution into $B = \frac{\mu_0 N I}{2r}$ gives $2\pi \times 10^{-4}$ T, which is approximately $6.28 \times 10^{-4}$ T.

Question 78 (Field at the center)

   Question: Magnetic field intensity at the centre of the coil of 50 turns, radius 0.5 m and carrying a current of 2 A, is

   Options: 

       A. $3 \times 10^{-5}$ T

       B. $1.25 \times 10^{-4}$ T

       C. $0.5 \times 10^{-5}$ T

       D. $4 \times 10^6$ T

   Correct Answer: B

   Year: 1999

   Solution: $B = \frac{\mu_0 (N i)}{2 r} = \frac{4 \pi \times 10^{-7} \times 50 \times 2}{2 \times 0.5} = 1.256 \times 10^{-4}$ T.

   Step Solution:

    1.  Identify given values: Number of turns $N = 50$, radius $r = 0.5$ m, and current $I = 2$ A.

    2.  State the formula: Use the magnetic field formula for the center of a circular coil: $B = \frac{\mu_0 NI}{2r}$.

    3.  Substitute values: $B = \frac{4\pi \times 10^{-7} \times 50 \times 2}{2 \times 0.5}$.

    4.  Simplify denominator and numerator: The denominator $2 \times 0.5 = 1$. The numerator becomes $4\pi \times 10^{-7} \times 100 = 4\pi \times 10^{-5}$.

    5.  Final calculation: $B = 4 \times 3.14 \times 10^{-5} = 12.56 \times 10^{-5}$ T, which is approximately $1.25 \times 10^{-4}$ T.

   Difficulty Level: Easy

   Concept Name: Magnetic field at the center of a circular coil.

   Short cut solution: Since the denominator $2r = 1$, the field is simply $\mu_0 \times (N \times I) = (4\pi \times 10^{-7}) \times 100 = 4\pi \times 10^{-5} \approx \mathbf{1.25 \times 10^{-4} \text{ T}}$.

Question 82 (Field at the center)

   Question: A coil of one turn is made of a wire of certain length and then from the same length a coil of two turns is made. If the same current is passed in both the cases, then the ratio of the magnetic inductions at their centres will be

   Options: 

       A. 4: 1

       B. 1: 4

       C. 2: 1

       D. 1: 2

   Correct Answer: B

   Year: 1998

   Solution: Magnetic field at the centre of the coil, $B = \frac{\mu_0}{2 \pi} \frac{NI}{a}$. Let $l$ be the length of the wire, then $B_1 = \frac{\mu_0}{2 \pi} \cdot \frac{1 \times I}{l/2\pi} = \frac{\mu_0 I}{l}$. $B_2 = \frac{\mu_0}{2 \pi} \cdot \frac{2 \times I}{l/4\pi} = \frac{4 \mu_0 I}{l}$. Therefore, $\frac{B_1}{B_2} = \frac{1}{4}$ or, $B_1 : B_2 = 1 : 4$.

   Step Solution:

    1.  Relate length and radius: For a fixed length $L$, one turn gives $L = 2\pi r_1 \implies r_1 = \frac{L}{2\pi}$. Two turns give $L = 2(2\pi r_2) \implies r_2 = \frac{L}{4\pi}$.

    2.  Calculate field for 1 turn: $B_1 = \frac{\mu_0 (1) I}{2 r_1} = \frac{\mu_0 I}{2(L/2\pi)} = \frac{\mu_0 \pi I}{L}$.

    3.  Calculate field for 2 turns: $B_2 = \frac{\mu_0 (2) I}{2 r_2} = \frac{2 \mu_0 I}{2(L/4\pi)} = \frac{4 \mu_0 \pi I}{L}$.

    4.  Set up the ratio: $\frac{B_1}{B_2} = \frac{\mu_0 \pi I / L}{4 \mu_0 \pi I / L}$.

    5.  Simplify: The ratio is $\frac{1}{4}$ or $1:4$.

   Difficulty Level: Medium

   Concept Name: Magnetic field at the center of a circular coil with variable turns.

   Short cut solution: For a fixed length of wire bent into $N$ turns, the magnetic field at the center is $B \propto N^2$. Since $N$ increases from 1 to 2, the field becomes $2^2 = 4$ times larger. Thus, the ratio is $1:4$.

Question 14 (Field on the axis)

   Question: The magnetic field on the axis of a circular loop of radius 100 cm carrying current $\mathbf{I = \sqrt{2} A}$, at point 1m away from the centre of the loop is given by:

   Options: 

    A. $6.28 \times 10^{-4} \text{ T}$

    B. $3.14 \times 10^{-7} \text{ T}$

    C. $6.28 \times 10^{-7} \text{ T}$

    D. $3.14 \times 10^{-4} \text{ T}$

   Correct Answer: B

   Year: NEET Re-2022

   Solution (as given in the source): 

    $= \frac{4\pi \times 10^{-7} \times \sqrt{2} \times 1^2}{2 (1^2 + 1^2)^{3/2}}$

    $= \frac{4\pi \times 10^{-7} \times \sqrt{2}}{2 \times 2\sqrt{2}} = \pi \times 10^{-7} \text{ T}$

    $= 3.14 \times 10^{-7} \text{ T}$

   Step Solution:

    1.  Identify Given Values: Radius $R = 100 \text{ cm} = 1 \text{ m}$, distance from center $x = 1 \text{ m}$, and current $I = \sqrt{2} \text{ A}$.

    2.  State the Axial Field Formula: The magnetic field on the axis is $B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.

    3.  Substitute the Values: $B = \frac{(4\pi \times 10^{-7}) \times \sqrt{2} \times (1)^2}{2(1^2 + 1^2)^{3/2}}$.

    4.  Simplify the Denominator: Since $R^2 + x^2 = 1 + 1 = 2$, the term $(2)^{3/2} = 2\sqrt{2}$. The denominator becomes $2 \times 2\sqrt{2} = 4\sqrt{2}$.

    5.  Calculate Final Result: $B = \frac{4\pi \times 10^{-7} \times \sqrt{2}}{4\sqrt{2}} = \pi \times 10^{-7} \text{ T} \approx 3.14 \times 10^{-7} \text{ T}$.

   Difficulty Level: Medium

   Concept Name: Magnetic field on the axis of a circular loop.

   Short cut solution: When the distance from the center $x$ equals the radius $R$, the formula simplifies to $B_{axis} = \frac{B_{center}}{2\sqrt{2}}$. Here, $B_{center} = \frac{\mu_0 I}{2R} = \frac{4\pi \times 10^{-7} \times \sqrt{2}}{2 \times 1} = 2\pi\sqrt{2} \times 10^{-7}$. Dividing by $2\sqrt{2}$ gives $\pi \times 10^{-7} \text{ T}$, which is $3.14 \times 10^{-7} \text{ T}$.

Question 7 (Field from bent/semi-circular wires)

   Question: A very long conducting wire is bent in a semi-circular shape from A to B as shown in figure. The magnetic field at point P for steady current configuration is given by.

   Options: 

       A. $\frac { \mu _ { 0 } i } { 4 R }$ pointed away from the page.

       B. $\frac { \mu _ { 0 } i } { 4 R } \Big [ 1 - \frac { 2 } { \pi } \Big ]$ pointed away from page.

       C. $\frac { \mu _ { 0 } i } { 4 R } \Big [ 1 - \frac { 2 } { \pi } \Big ]$ pointed into the page.

       D. $\frac { \mu _ { 0 } i } { 4 R }$ pointed into the page.

   Correct Answer: B.

   Year: NEET 2023.

   Solution:

$B _ { p }$ due to wire $1 = \frac { \mu _ { 0 } i } { 4 \pi R } \otimes$; $B _ { p }$ due to wire $3 = \frac { \mu _ { 0 } i } { 4 \pi R } \otimes$; $B _ { p }$ due to wire $2 = \frac { \mu _ { 0 } i } { 4 R } \odot$. $B _ {\text {net}} = - \frac {\mu_ {0} i}{2 \pi R} + \frac {\mu_ {0} i}{4 R} = \frac {\mu_ {0} i}{4 R} \left[ 1 - \frac {2}{\pi} \right]$ pointed away from page.

   Step Solution:

    1.  Analyze segments: The setup consists of two semi-infinite straight wires (1 and 3) and one semi-circular arc (2) of radius $R$.

    2.  Calculate field from straight wires: At point $P$, each semi-infinite wire produces a magnetic field $B = \frac{\mu_0 i}{4\pi R}$ directed into the page ($\otimes$).

    3.  Calculate field from semi-circle: The semi-circular arc produces a magnetic field $B = \frac{\mu_0 i}{4R}$ at the center, directed away from the page ($\odot$).

    4.  Vector Summation: $B_{net} = \frac{\mu_0 i}{4R} - (\frac{\mu_0 i}{4\pi R} + \frac{\mu_0 i}{4\pi R}) = \frac{\mu_0 i}{4R} - \frac{\mu_0 i}{2\pi R}$.

    5.  Simplify: Factor out $\frac{\mu_0 i}{4R}$ to get $B_{net} = \frac{\mu_0 i}{4R} [1 - \frac{2}{\pi}]$.

   Difficulty Level: Medium

   Concept Name: Superposition principle and magnetic fields of semi-infinite wires and circular arcs.

   Short cut solution: Since the field from the arc ($\propto 1$) is stronger than the combined field of the wires ($\propto 2/\pi \approx 0.63$), the net field must be outward (away from page), leaving only option B as a viable choice.

Question 33 (Field from bent/semi-circular wires)

   Question: A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to X-axis while semi circular portion of radius R is lying in Y-Z plane. Magnetic field at point O is.

   Options: 

       A. $\vec { \mathrm { B } } = - \frac { \mu _ { 0 } } { 4 \pi } \frac { \mathrm { I } } { \mathrm { R } } ( \pi \hat{i} + 2 \hat{k} )$.

       B. $\vec { \mathrm { B } } = \frac { \mu _ { 0 } } { 4 \pi } \frac { \mathrm { I } } { \mathrm { R } } ( \pi \hat{i} - 2 \hat{j} )$.

       C. $\vec { \mathrm { B } } = \frac { \mu _ { 0 } } { 4 \pi } \frac { \mathrm { I } } { \mathrm { R } } ( \pi \hat{i} + 2 \hat{k} )$.

       D. $\vec { \mathrm { B } } = - \frac { \mu _ { 0 } } { 4 \pi } \frac { \mathrm { I } } { \mathrm { R } } ( \pi \hat{i} - 2 \hat{k} )$.

   Correct Answer: A.

   Year: 2015.

   Solution: Wires 1 and 3 are semi-infinite: $\vec{B}_1 = \vec{B}_3 = -\frac{\mu_0 I}{4\pi R} \hat{k}$. Semi-circular arc in Y-Z plane: $\vec{B}_2 = -\frac{\mu_0 I}{4R} \hat{i}$. Net field: $\vec{B} = \vec{B}_1 + \vec{B}_2 + \vec{B}_3 = -\frac{\mu_0 I}{4\pi R} (2\hat{k} + \pi\hat{i})$.

   Step Solution:

    1.  Identify components: The wire has two semi-infinite straight sections along the x-axis and a semi-circle in the y-z plane.

    2.  Field of straight wires: For segments 1 and 3, the field at the origin is $\vec{B}_1 = \vec{B}_3 = \frac{\mu_0 I}{4\pi R} (-\hat{k})$.

    3.  Field of semi-circle: For the arc in the y-z plane, the field at the center is $\vec{B}_2 = \frac{\mu_0 I}{4R} (-\hat{i})$.

    4.  Vector Addition: $\vec{B}_{total} = \vec{B}_1 + \vec{B}_2 + \vec{B}_3 = -2(\frac{\mu_0 I}{4\pi R} \hat{k}) - \frac{\mu_0 I \pi}{4\pi R} \hat{i}$.

    5.  Factor Result: $\vec{B} = -\frac{\mu_0 I}{4\pi R} (\pi\hat{i} + 2\hat{k})$.

   Difficulty Level: Hard

   Concept Name: Vector superposition of magnetic fields in 3D using Biot-Savart Law components.

   Short cut solution: Note that the field from the arc in the Y-Z plane must be along the X-axis ($-\hat{i}$ direction), and the straight wires along the X-axis must produce a field in the Z-direction ($-\hat{k}$); only option A matches this vector orientation.