Mutual Inductance Video Lessons

Concept

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.

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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.