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

Step 1: Understanding the Concept:
The magnetic field at the center of a circular current loop is $B = \frac{\mu_0 I}{2R}$. For a semicircle, it is half of this value. For multiple segments, the total field is the vector sum. Straight segments passing through the center contribute zero field.
Step 2: Key Formula or Approach:
1. Magnetic field of semicircle: $B_{semi} = \frac{\mu_0 I}{4R}$
2. Use Right Hand Thumb Rule to determine directions.
Step 3: Detailed Explanation:
1. For the inner semicircle of radius $R_2$: The current flows in a direction that creates a magnetic field directed out of the page (assume CCW).
2. For the outer semicircle of radius $R_1$: The current flows in the opposite direction (CW), creating a field directed into the page.
3. The magnitudes are $B_2 = \frac{\mu_0 I}{4R_2}$ and $B_1 = \frac{\mu_0 I}{4R_1}$.
4. Since $R_2<R_1$, the field $B_2$ is stronger. Net field magnitude:
\[ B_{net} = B_2 - B_1 = \frac{\mu_0 I}{4R_2} - \frac{\mu_0 I}{4R_1} = \frac{\mu_0 I}{4} \left[ \frac{1}{R_2} - \frac{1}{R_1} \right] \]
Step 4: Final Answer:
The magnitude of the magnetic field is $\frac{\mu_0 I}{4} \left[ \frac{1}{R_2} - \frac{1}{R_1} \right]$.