BITSAT 2013 Paper

The magnetic field at C due to first conductor is
B1 =

µ0

2π

I

3d∕2

(since, point C is separated by d +

d

2

=

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
B2 =

µ0

2π

10

d∕2

(since, point C is separated by

d

2

from 2nd conductor)
The direction of field is perpendicular to the plane of paper and directed inwards. Since, direction of B1 and B2 at point C is in opposite direction and the magnetic field at C is zero, therefore,
B1 = B2

µ0

2π

I

3d∕2

=

µ0

2π

10

d∕2

On solving I = 30.0 A