Source And Effects Of Magnetic Field | Physics JEE previous year questions with Solutions

Q 1 :

Two insulated circular loops A and B of radius 'a' carrying a current of 'I' in the anti-clockwise direction as shown in figure. The magnitude of the magnetic induction at the centre will be ______. [2024]

$\frac{\sqrt{2} μ_{0} I}{2 a}$
$\frac{μ_{0} I}{2 a}$
$\frac{\sqrt{2} μ_{0} I}{a}$
$\frac{2 μ_{0} I}{a}$

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(1)

${\vec{B}}_{A} = \frac{μ_{0} I}{2 a}$

${\vec{B}}_{B} = \frac{μ_{0} I}{2 a}$ $∵$ ${\vec{B}}_{net} = \frac{\sqrt{2} μ_{0} I}{2 a}$

Q 2 :

The magnetic field at the centre of a wire loop formed by two semicircular wires of radii $R_{1} = 2 π m$ and $R_{2} = 4 π m$ , carrying current $I = 4 A$ as per figure given below is $α \times 10^{- 7} T$ . The value of $α$ is ______. (Centre O is common for all segments) [2024]

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(3)

$\frac{μ_{0} i}{2 R_{2}} (\frac{π}{2 π}) \otimes + \frac{μ_{0} i}{2 R_{1}} (\frac{π}{2 π}) \otimes$

$(\frac{μ_{0} i}{4 R_{2}} + \frac{μ_{0} i}{4 R_{1}}) \otimes$

$\frac{4 π \times 10^{- 7} \times 4}{4 \times 4 π} + \frac{4 π \times 10^{- 7} \times 4}{4 \times 2 π}$

$= 3 \times 10^{- 7} = α \times 10^{- 7}$

$α = 3$

Q 3 :

Two circular coils P and Q of 100 turns each have same radius of $π$ cm. The currents in P and R are 1A and 2A respectively. P and Q are placed with their planes mutually perpendicular with their centers coincide. The resultant magnetic field induction at the center of the coils is $\sqrt{x} m T,$ where $x =$ ____ .

$[Use μ_{0} = 4 π \times 10^{- 7} {TmA}^{- 1}]$ [2024]

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(20)

$B_{P} = \frac{μ_{0} N i_{1}}{2 r} = \frac{μ_{0} \times 1 \times 100}{2 π} = 2 \times 10^{- 3} T$

$B_{Q} = \frac{μ_{0} N i_{2}}{2 r} = \frac{μ_{0} \times 2 \times 100}{2 π} = 4 \times 10^{- 3} T$

$B_{net} = \sqrt{B_{P}^{2} + B_{Q}^{2}} = \sqrt{20} mT$

$x = 20$

Q 4 :

An infinite wire has a circular bend of radius a, and carrying a current $I$ as shown in figure.

The magnitude of magnetic field at the origin O of the arc is given by: [2025]

$\frac{μ_{0}}{4 π} \frac{I}{a} [\frac{π}{2} + 1]$
$\frac{μ_{0}}{4 π} \frac{I}{a} [\frac{3 π}{2} + 1]$
$\frac{μ_{0}}{2 π} \frac{I}{a} [\frac{π}{2} + 2]$
$\frac{μ_{0}}{4 π} \frac{I}{a} [\frac{3 π}{2} + 2]$

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(2)

$B_{1} = \frac{μ_{0} I}{4 π a} \otimes$

$B_{2} = \frac{μ_{0}}{4 π} \frac{I}{a} (\frac{3 π}{2}) \otimes$

$B_{3} = 0$

$B = \frac{μ_{0}}{4 π} \frac{I}{a} (1 + \frac{3 π}{2}) \otimes$

Q 5 :

Consider a long straight wire of a circular cross-section (radius a) carrying a steady current $I$ . The current is uniformly distributed across this cross-section. The distances from the centre of the wire's cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be [2025]

$[\frac{a}{4}, \frac{a}{2}]$
$[\frac{a}{2}, 2 a]$
$[\frac{a}{2}, 3 a]$
$[\frac{a}{4}, 2 a]$

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Maximum possible magnetic field on the surface

$B_{\max} = \frac{μ_{0} I}{2 π a} \Rightarrow \frac{B_{\max}}{2} = \frac{μ_{0} I}{4 π a}$

It can be obtained inside as well as outside the wire.

For inside, $\frac{μ_{0} I}{4 π a} = \frac{μ_{0} I r}{2 π a^{2}} \Rightarrow r = \frac{a}{2}$

For outside, $\frac{μ_{0} I}{4 π a} = \frac{μ_{0} I}{2 π r} \Rightarrow r = 2 a$

Correct answer $[\frac{a}{2}, 2 a]$

Q 6 :

Let $B_{1}$ be the magnitude of magnetic field at center of a circular coil of radius R carrying current $I$ . Let $B_{2}$ be the magnitude of magnetic field at an axial distance 'x' from the centre. For $x : R = 3 : 4, \frac{B_{2}}{B_{1}}$ is: [2025]

4 : 5
16 : 25
64 : 125
25 : 16

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(3)

The magnitude of magnetic field at centre of a circular coil, $B_{1} = \frac{μ_{0} i}{2 R}$

The magnitude of magnetic field at an axial distance $x = \frac{3 R}{4}$ from the centre.

$B_{2} = \frac{μ_{0} i R^{2}}{2 {(R^{2} + x^{2})}^{3 / 2}} = \frac{μ_{0} i R^{2}}{2 {(\frac{5 R}{4})}^{3}} = \frac{64}{125} (\frac{μ_{0} i}{2 R})$

$\Rightarrow \frac{B_{2}}{B_{1}} = \frac{64}{125}$

Q 7 :

A loop ABCDA, carrying current $I$ = 12A, is placed in a plane, consists of two semi-circular segments of radius $R_{1} = 6 π m$ and $R_{2} = 4 π m$ . The magnitude of the resultant magnetic field at center O is $k \times 10^{- 7} T$ . The value of k is _______ (Given $μ_{0} = 4 π \times 10^{- 7} T {mA}^{- 1}$ ) [2025]

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(1)

Magnetic field due to AB and CD = 0

$B = \frac{μ_{0} I}{2 r} (\frac{θ}{2 π})$ for an arc

For semi-circule $B = \frac{μ_{0} I}{4 r}$

$B_{net} = \frac{μ_{0} I}{4 {4 π}} - \frac{μ_{0} I}{4 {6 π}}$

$= \frac{μ_{0} I}{4 π} {\frac{1}{4} - \frac{1}{6}}$

$= 10^{- 7} \times 12 \times \frac{2}{24} = 10^{- 7} T \Rightarrow k = 1$

Q 8 :

Match the Column – I with Column – II: [2023]

[IMAGE 228]

Choose the correct answer from the options given below:

A – I; B – III; C – IV; D – II
A – III; B – IV; C – I; D – II
A – II; B – I; C – IV; D – III
A – III; B – I; C – IV; D – II

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(4)

$A \to B_{0} = \frac{μ_{0}}{4 π} \frac{i}{r} ⊙ + \frac{μ_{0} i}{2 r} \otimes + \frac{μ_{0}}{4 π} \frac{i}{r} ⊙$

$= \frac{μ_{0}}{2 π} \frac{i}{r} ⊙ + \frac{μ_{0} i}{2 r} \otimes$

$B_{0} = \frac{μ_{0} i}{2 π r} (π - 1) \otimes$

$B \to B_{0} = \frac{μ_{0}}{4 π} \frac{i}{r} ⊙ + \frac{μ_{0}}{4 π} \frac{i}{r} π ⊙ + \frac{μ_{0}}{4 π} \frac{i}{r} π ⊙$

$= (\frac{2 μ_{0}}{4 π} \frac{i}{r} + \frac{μ_{0}}{4 π} \frac{i}{r} π) ⊙$

$B_{0} = \frac{μ_{0} i}{4 π r} (2 + π) ⊙$

$C \to B_{0} = \frac{μ_{0}}{4 π} \frac{i}{r} \otimes + \frac{μ_{0}}{4 π} \frac{i}{r} π \otimes + 0$

$B_{0} = \frac{μ_{0}}{4 π} \frac{i}{r} (1 + π) \otimes$

$D \to B_{0} = \frac{μ_{0}}{4 π} \frac{i}{r} π = \frac{μ_{0}}{4 π} \frac{i}{r} ⊙$

$A - III; B - I; C - IV; D - II$

Q 9 :

A single current carrying loop of wire carrying current $I$ flowing in anticlockwise direction seen from +ve $z$ direction and lying in $x y$ plane as shown in figure. The plot of $\hat{j}$ component of magnetic field $(B_{y})$ at a distance $' a'$ (less than radius of the coil) and on $y z$ plane vs $z$ coordinate looks like [2023]

[IMAGE 229]

[IMAGE 230]
[IMAGE 231]
[IMAGE 232]
[IMAGE 233]

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(3)

[IMAGE 234]

$B_{y} = 0 in the plane of the coil$

$B_{y} is opposite to each other at - z and + z positions.$

Q 10 :

The electric current in a circular coil of four turns produces a magnetic induction 32 T at its centre. The coil is unwound and is rewound into a circular coil of single turn. The magnetic induction at the centre of the coil by the same current will be [2023]

8 T
4 T
2 T
16 T

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(3)

$B = \frac{μ_{0} i}{2 R} \times 4 and B' = \frac{μ_{0} i}{2 R'}$

$For R' = 4 R \Rightarrow B' = \frac{μ_{0} i}{8 R}$

$\frac{B'}{B} = \frac{1}{16} \Rightarrow B' = 2 T$

Topic Question Set

Q 1 :

Two insulated circular loops A and B of radius 'a' carrying a current of 'I' in the anti-clockwise direction as shown in figure. The magnitude of the magnetic induction at the centre will be ______. [2024]

Q 2 :

The magnetic field at the centre of a wire loop formed by two semicircular wires of radii $R_{1} = 2 π m$ and $R_{2} = 4 π m$ , carrying current $I = 4 A$ as per figure given below is $α \times 10^{- 7} T$ . The value of $α$ is ______. (Centre O is common for all segments) [2024]

0%

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Q 4 :

An infinite wire has a circular bend of radius a, and carrying a current $I$ as shown in figure.

The magnitude of magnetic field at the origin O of the arc is given by: [2025]

Q 6 :

Let $B_{1}$ be the magnitude of magnetic field at center of a circular coil of radius R carrying current $I$ . Let $B_{2}$ be the magnitude of magnetic field at an axial distance 'x' from the centre. For $x : R = 3 : 4, \frac{B_{2}}{B_{1}}$ is: [2025]

0%

Q 8 :

Match the Column – I with Column – II: [2023]

[IMAGE 228]

Choose the correct answer from the options given below:

Q 10 :

The electric current in a circular coil of four turns produces a magnetic induction 32 T at its centre. The coil is unwound and is rewound into a circular coil of single turn. The magnetic induction at the centre of the coil by the same current will be [2023]

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