The wire loop PQRSP formed by joining two semicircular wires of radii R1 and R2 carries a current I as shown in the figure.
The magnitude of the magnetic induction at the centre C is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (μ0I ÷ 4) (1/R1 − 1/R2)
Each semicircular arc gives half the field of a full circular loop at its centre. The straight parts PQ and RS lie along lines through C, so for them \( d\vec{l} \) is parallel to \( \vec{r} \) and they give no field at C.
\( \displaystyle B_{\mathrm{arc}} = \frac{1}{2} \times \frac{\mu_0 I}{2R} = \frac{\mu_0 I}{4R} \)
From the figure, the current goes round the outer arc (radius R2) clockwise, from S over the top to P, and round the inner arc (radius R1) anticlockwise, from Q over the top to R. By the right-hand thumb rule the two fields at C are in opposite directions: into the page for the outer arc and out of the page for the inner arc.
\( \displaystyle B_1 = \frac{\mu_0 I}{4R_1} \qquad B_2 = \frac{\mu_0 I}{4R_2} \)
Since \( R_1 \lt R_2 \), \( B_1 \) is the larger, and the resultant field at C is
\( \displaystyle B = B_1 - B_2 = \frac{\mu_0 I}{4}\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)