# Problem: A solenoidal coil with 25 turns of wire is wound tightly around another coil with 320 turns. The inner solenoid is 23.0cm long and has a diameter of 2.00 cm. At a certain time, the current in the inner solenoid is 0.150A and is increasing at a rate of 1600A/sPart A. For this time, calculate the average magnetic flux through each turn of the inner solenoid.Part B. For this time, calculate the mutual inductance of the two solenoids;Part C. For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.

###### FREE Expert Solution

This problem requires us to remember the concept of magnetic flux in a solenoid, given by:

$\overline{){\mathbf{\varphi }}{\mathbf{=}}{\mathbf{B}}{\mathbf{A}}}$, where is the magnetic field and A is the area.

Magnetic field due to solenoid:

$\overline{){\mathbf{B}}{\mathbf{=}}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{iN}}{\mathbf{L}}}$, where i is current, N is the number of turns, and L is the length of the solenoid.

88% (53 ratings) ###### Problem Details

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 320 turns. The inner solenoid is 23.0cm long and has a diameter of 2.00 cm. At a certain time, the current in the inner solenoid is 0.150A and is increasing at a rate of 1600A/s

Part A. For this time, calculate the average magnetic flux through each turn of the inner solenoid.

Part B. For this time, calculate the mutual inductance of the two solenoids;

Part C. For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.