Question

Two circular loops carry identical currents, but the radius of one loop is twice that of the other.

Answer 1

The magnetic field at the center of the 2nd loop is half the magnetic field at the center of the 1st loop.

Explanation:

The magnetic field strength at the center of a current-carrying circular loop is

$B=\frac{\mu_0 I}{2R}$

where

$\mu_0$ is the vacuum permeability

I is the current

R is the radius of the loop

For the first circular loop, we have

$B_1 = \frac{\mu_0 I_1}{2R_1}$

The second loop has

$I_2 = I_1$ (same current)

$R_2 = 2R_1$ (twice the radius)

So, the magnetic field at the centre of the second loop is

$B_2 = \frac{\mu_0 I_2}{2R_2}=\frac{1}{2}(\frac{\mu_0 I_1}{2R_1})=\frac{B_1}{2}$

So, the magnetic field at the center of the 2nd loop is half the magnetic field at the center of the 1st loop.

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