Question

Two equal-length solenoids are constructed using copper wires. The first solenoid has 400 loops and has a diameter of 4cm. The second solenoid has 1200 loops and a diameter of 6cm. The currents through each solenoid are equal.

Show that the ratio of the magnetic fields of the two solenoids, B2/B1, is equal to 3.

Answer 1

Magnetic field inside a solenoid is given by the Expression,

B = μ*N*I/L

where B is the magnetic field inside solenoid,μ is permiability,N is the total number of loops and I is the current in the solenoid having length L.

For first solenoid,

B1 = μ*400*I/L -------->(1)

For second solenoid

B2 = μ*1200*I/L --------> (2)

as both having same length L and current I.

dividing equation (2) by equation (1),

B2/B1 = (μ*1200*I/L)/(μ*400*I/L)

B2/B1 = 1200/400

B2/B1 = 3

Hence the ratio of the magnetic field inside the solenoids is 3.