Question

The wire loop $PQRSP$ formed by joining two semicircular wire of radii $R_1$ and $R_2$ carries a current $I$ as shown. The magnitude of the magnetic field at the centre ' O ' is

• A. $\frac{\mu_0 I}{4} \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$
• B. $\frac{\mu_0 I}{4} \left[ \frac{1}{R_2} - \frac{1}{R_1} \right]$
• C. $\frac{\mu_0 I}{2\pi} \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$
• D. $\frac{\mu_0 I}{2\pi} \left[ \frac{1}{R_2} - \frac{1}{R_1} \right]$

Hint:

At the centre of a semicircle: \[ B=\frac{\mu_0 I}{4R} \] Straight radial segments through the centre contribute zero magnetic field there.

Answer 1

The Correct Option is A

Concept:
Magnetic field at the centre due to a semicircular current carrying arc of radius \(R\) is: \[ B=\frac{\mu_0 I}{4R} \] The two straight connecting segments pass through the centre line, so their contribution at the centre is zero. ip

Step 1: Write field due to each semicircle.
For the smaller semicircle of radius \(R_1\): \[ B_1=\frac{\mu_0 I}{4R_1} \] For the larger semicircle of radius \(R_2\): \[ B_2=\frac{\mu_0 I}{4R_2} \] ip

Step 2: Combine the two fields.
From the figure, the currents in the two semicircular parts produce magnetic fields at the centre in opposite directions. So net magnitude is: \[ B=B_1-B_2 \] \[ B=\frac{\mu_0 I}{4}\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \] ip Hence, the correct answer is:
\[ \boxed{(A)\ \frac{\mu_0 I}{4}\left[\frac{1}{R_1}-\frac{1}{R_2}\right]} \]