Question

A solenoidal coil with 30 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 24.0 cm long and has a diameter of 2.30 cm . At a certain time, the current in the inner solenoid is 0.150 A and is increasing at a rate of 1600 A/s . For this time, calculate the average magnetic flux through each turn of the inner solenoid.For this time, calculate the mutual inductance of the two solenoids;For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.

Answer 1

Given:

Number of turns in the inner coil N₁ = 300 turns

Number of tuns in the outer coil N₂ = 30 turns

Length of the inner solenoid l = 24.0 cm = 0.24 m

Radius of the inner coil:

Current in the inner solenoid I₁ = 0.15 A

Rate of increase of current

Solution:

Area of cross-section of the inner solenoid:

Magnetic field through the inner solenoid:

Magnetic flux through each turn of the inner solenoid is given by:

As magnetic field vector and area directed along the axis.

The magnetic flux through each turn of the outer coil is also equal .

Therefore,

The mutual inductance of the two solenoids is given by:

The emf induced in the outer solenoid is given by: