A vertical circular coil of radius 0.1 ~m and having 10 turns carries a steady current. When the plane of the coil is normal to the magnetic meridian, a neutral point is observed at the centre of the coil. If B_ H =0.314 10⁻⁴ ~T , then the current in the coil is

Step-by-step solution

Given, radius of the coil =0.1 ~m No. of turns in the coil =10 Horizontal component of magnetic field, B_ H =0.314 10⁻⁴ ~T The magnetic field at the centre of current carrying coil is given by the formula, B= ₀ n I 2 R According to question, magnetic field due to the coil obtains neutral point, in earth's magnetic field. Hence, horizontal component of magnetic field, aligned & B_ H = ₀ n I 2 R Thus, I &= 2 R B_ H ₀ n I &= 2(0.1) (0.314 10⁻⁴ ) 10 4 (3.142) 10⁻⁷ =0.5 ~A aligned