A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of magnetic field will be equal to

Step-by-step solution

Magnetic field due to long solenoid along its axis is given as, B = μ 0 n I , where n is the number of turns in the solenoid per unit length, I is current in the solenoid and μ 0 is a constant called the permeability of free space. ⇒ B 1   B 2 = n 1 I 1 n 2 I 2 According to the question, n 2 = n 1 2   and   I 2 = 2 I 1 ⇒ B B 2 = n 1 n 1 2 × I 1 2 I 1 ⇒ B 2 = B ; means, the magnetic field remains the same.