The given figure shows a square loop made from a uniform wire. If a battery is connected between the point A and C, the magnetic field at the centre O of the square would be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Zero
The current I entering at A divides between the two paths A-B-C and A-D-C. The wire is uniform and both paths have two sides, so their resistances are equal and the current divides equally, \( I/2 \) in each.
The field at the centre O is the sum of the fields of the four sides, from the Biot-Savart law. Side AB and side AD carry the same current \( I/2 \) and are the same distance from O, so they give fields of equal magnitude. Their directions are opposite (AB drives the current one way round O, AD the other), so the two fields cancel. In the same way the fields of BC and DC cancel.
\( B_{\mathrm{O}} = 0 \)