The wire loop PQRSP formed by joining two semicircular wires of radii \( R_1 \) & \( R_2 \) carries a current I as shown in the fig. The magnitude of magnetic field at the centre 'C' is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{\mu_0 I}{4}\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \)
By the Biot-Savart law, a full circular loop of radius \( R \) gives \( \mu_0 I/(2R) \) at its centre, so a semicircular arc gives half of that.
\( \displaystyle B_{\mathrm{arc}} = \frac{\mu_0 I}{4R} \)
The straight parts PQ and RS lie along a line through C. For every element of them \( d\vec{l} \) is parallel to the line joining it to C, so \( d\vec{l}\times\hat{r} = 0 \) and they give no field at C.
From the arrows in the figure, the current goes round the inner arc QR (radius \( R_1 \)) and the outer arc SP (radius \( R_2 \)) in opposite senses. So their fields at C point in opposite directions (one out of the page, the other into it) and are subtracted.
\( \displaystyle B = \frac{\mu_0 I}{4R_1} - \frac{\mu_0 I}{4R_2} = \frac{\mu_0 I}{4}\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \)
Since \( R_1 \lt R_2 \), this is positive, and it is the magnitude of the field.