A copper wire with a resistance of 0.01 ohm per metre is utilized to wind a 400 -turn solenoid having a radius of 1.0 cm and a length of 20 cm . Determine the emf of a battery which, when connected across the solenoid, will produce a magnetic field of 1.0 10⁻² T near the centre of the solenoid.

Step-by-step solution

The total length of the wire used to wind the solenoid is given by: l = N 2 r = 400 2 0.01 = 8 m The total resistance of the solenoid is: R = l = 8 0.01 = 0.08 The magnetic field near the centre of a long solenoid is: B = ₀ n I = ₀ ( N L ) I Substituting the given values to find the current I : 1.0 10⁻² = (4 10⁻⁷) ( 400 0.2 ) I 1.0 10⁻² = 8 10⁻⁴ I I = 1.0 10⁻² 8 10⁻⁴ = 100 8 = 12.5 A The emf of the battery required is: E = I R = ( 12.5 ) 0.08 = 1.0 V