A rectangular coil comprising 100 turns has a length of 5 cm and a width of 4 cm . It is positioned such that its plane is parallel to a uniform magnetic field, and a current of 2 A is passed through the coil. Determine the magnitude of the magnetic field B if the torque acting on the coil is 0.2 N m ⁻¹ .

Step-by-step solution

The torque acting on a current-carrying coil in a uniform magnetic field is given by: = N I A B Given that the plane of the coil is parallel to the magnetic field, the angle between the area vector and the magnetic field is = 90^ . Substituting the given values: N = 100 I = 2 A A = 5 cm 4 cm = 20 cm ^2 = 20 10⁻⁴ m ^2 = 0.2 N m We get: 0.2 = 100 2 20 10⁻⁴ B (90^ ) 0.2 = 4000 10⁻⁴ B 0.2 = 0.4 B B = 0.2 0.4 = 0.5 T