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Magnetic Effects & Magnetism

JEE Main level, single correct, medium

73 of 276

Question 73 of 276

A wire $A$ , bent in the shape of an arc of a circle, carrying a current of $2A$ and having radius $2cm$ and another wire $B$ , also bent in the shape of an arc of a circle, carrying a current of $3A$ and having radius of $4\text{cm}$ , are placed as shown in the figure. The ratio of the magnetic fields due to the wires $\text{A}$ and $\text{B}$ at the common centre $\text{O}$ is:

Show answer and solution

Answer: (D) $6:5$

Given,
Current in the arc

A

{i}_{1}=2A

Current in the arc

B

{i}_{2}=3A

Radius of arc

A

{R}_{1}=2cm

Radius of arc

B

{R}_{2}=4cm

Magnetic field at the centre of circular current carrying arc is given by

{B}_{C}=\dfrac{{\mu }_{0}i}{4\pi R}\left(\theta \right)...(1)

Where,

\theta

is the angle subtended at centre.
Magnetic field at the centre due to arc

A

and

B

is given by

{B}_{A}=\dfrac{{\mu }_{0}{i}_{1}}{4{\pi R}_{1}}\left(2\pi -\dfrac{\pi }{2}\right)...\left(2\right)

{B}_{B}=\dfrac{{\mu }_{0}{i}_{2}}{4{\pi R}_{2}}\left(2\pi -\dfrac{\pi }{3}\right)...\left(3\right)

Dividing equation (2) by (1), we get

\Rightarrow \dfrac{{B}_{A}}{{B}_{B}}=\dfrac{{i}_{1}}{{i}_{2}}\times \dfrac{{R}_{2}\left(2\pi -\dfrac{\pi }{2}\right)}{{R}_{1}\left(2\pi -\dfrac{\pi }{3}\right)}

\Rightarrow \dfrac{{B}_{A}}{{B}_{B}}=\dfrac{2}{3}\times \dfrac{4}{2}\times \dfrac{3\pi }{2}\times \dfrac{3}{5\pi }

\Rightarrow \dfrac{{B}_{A}}{{B}_{B}}=\dfrac{6}{5}

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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