A circular loop of radius r carries a current i . How must a long,...

The magnetic field at the centre of the circular loop due to its own current $i$ is given by:

$B_1 = \dfrac{\mu_0 i}{2r}$

The magnetic field at the centre due to the long straight wire carrying current $4i$ at a distance $d$ is:

$B_2 = \dfrac{\mu_0 (4i)}{2\pi d}$

For the net magnetic field at the centre to be zero, the magnitudes of the magnetic fields must be equal and their directions must be opposite. Equating the magnitudes:

$\dfrac{\mu_0 i}{2r} = \dfrac{\mu_0 (4i)}{2\pi d}$

$d = \dfrac{4r}{\pi}$

For the fields to be opposite at the centre, the straight wire must produce a magnetic field in the opposite direction to that of the loop. By the right-hand thumb rule, if the straight wire is placed near the loop, its current must flow in the direction opposite to the current in the nearest part of the circular loop. This ensures that the magnetic field produced by the straight wire at the centre opposes the magnetic field produced by the loop.