BITSAT 2013 Paper

The magnetic field at C due to first conductor is $B_1$ = $\frac{\mu_0}{2\pi} \frac{I}{\frac{3d}{2}}$ (since, point C is separated by d + $\frac{d}{2}$ = $\frac{3d}{2}$ from 1st conductor). The direction of field is perpendicular to the plane of paper and directed outwards. The magnetic field at C due to second conductor is $B_2$ = $\frac{\mu_0}{2\pi} \frac{10}{\frac{d}{2}}$ (since, point C is separated by $\frac{d}{2}$ from 2nd conductor) The direction of field is perpendicular to the plane of paper and directed inwards. Since, direction of $B_1$ and $B_2$ at point C is in opposite direction and the magnetic field at C is zero, therefore, $B_1$ = $B_2$ $\frac{\mu_0}{2\pi} \frac{I}{\frac{3d}{2}}$ = $\frac{\mu_0}{2\pi} \frac{10}{\frac{d}{2}}$ On solving I = 30.0 A