JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Magnetic Effects of Current

The electric current in a circular coil of 2 turns produces a magnetic induction

B_{1}

at its centre. The coil is unwound and is rewound into a circular coil of 5 turns and the same current produces a magnetic induction

B_{2}

at its centre.

The ratio of

\frac{B _{2}}{B _{1}}

is :

Answer: (B)

Solution

The magnetic field at the center of a circular coil is given by:

B = \frac{N μ _{0} i}{2 R}

For the first coil with

N_{1} = 2

turns:

B_{1} = \frac{N _{1} μ _{0} i}{2 R _{1}}

When the coil is unwound and rewound into

N_{2} = 5

turns, the total wire length remains constant, so:

2 π R_{1} N_{1} = 2 π R_{2} N_{2} ⟹ R_{2} = \frac{N _{1} R _{1}}{N _{2}}

For the second coil:

B_{2} = \frac{N _{2} μ _{0} i}{2 R _{2}}

Taking the ratio:

\frac{B _{2}}{B _{1}} = \frac{N _{2}}{N _{1}} \times \frac{R _{1}}{R _{2}} = \frac{N _{2}}{N _{1}} \times \frac{N _{2}}{N _{1}} = (\frac{N _{2}}{N _{1}})^{2} = (\frac{5}{2})^{2} = \frac{25}{4}

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option