In the circuit ARB and ATB are semicircles in which current is flowing as shown in figure. The magnetic field at the center S of the circular loop will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Zero
The field at S is the sum of the fields of the straight leads and of the two semicircles, each found from the Biot-Savart law.
Straight leads PA and BQ: S lies on the line of both leads, so for every element \( d\vec{l}\times\vec{r} = 0 \). They give no field at S.Semicircles: the current \( I \) splits at A. The two semicircles have the same length and so the same resistance, so each carries \( I/2 \). A semicircle of radius \( r \) gives at its centre
\( \displaystyle B = \frac{\mu_0\,(I/2)}{4r} = \frac{\mu_0 I}{8r} \)
In ARB the current goes over the top from A to B (clockwise about S), and in ATB it goes along the bottom from A to B (anticlockwise about S). The two fields are equal and opposite (one into, one out of the page), so they cancel.
\( \displaystyle B_{S} = \frac{\mu_0 I}{8r} - \frac{\mu_0 I}{8r} = 0 \)