Question

The radius of the current carrying circular coil is doubled keeping the current passing through it the same. Then the ratio of the magnetic field produced at the centre of the coil before the doubling of the radius to the magnetic field after doubling of the radius is:

• A. 1 : 2
• B. 2 : 1
• C. 3 : 2
• D. 3 : 1

Hint:

The magnetic field at the centre of a coil is inversely proportional to the radius of the coil. Doubling the radius halves the magnetic fiel(D)

Answer 1

The Correct Option is B

The magnetic field at the centre of a current-carrying circular coil is given by the formula: \[ B = \frac{\mu_0 I}{2r} \] where: - \( B \) is the magnetic field, - \( \mu_0 \) is the permeability of free space, - \( I \) is the current, - \( r \) is the radius of the coil.
When the radius of the coil is doubled, the formula for the magnetic field becomes: \[ B' = \frac{\mu_0 I}{2(2r)} = \frac{\mu_0 I}{4r} \] Thus, the new magnetic field \( B' \) is half of the original magnetic field \( B \).
The ratio of the magnetic field before and after doubling the radius is: \[ \frac{B}{B'} = \frac{\frac{\mu_0 I}{2r}}{\frac{\mu_0 I}{4r}} = \frac{4r}{2r} = 2 \] Therefore, the ratio of the magnetic field before doubling the radius to the magnetic field after doubling the radius is \( 2 : 1 \).