221k views
5 votes
suppose a 50 turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. the coil originally has an area of 0.375 m². it is squished to have no area in 0.175 s. What is the direction and magnitude of the induced emf if the uniform magnetic field has a strength of 1.50 T?

User Rui Yang
by
8.1k points

1 Answer

4 votes

Final answer:

The induced emf in the coil is -31.25 V. The negative sign indicates that the induced emf opposes the change in magnetic flux. The direction of the induced current would generate a magnetic field directed out of the page, opposing the initial magnetic field.

Step-by-step explanation:

The induced emf can be found using Faraday's Law of electromagnetic induction, which states that the induced emf is equal to the negative rate of change of magnetic flux through the coil.

The formula for calculating the induced emf is given by:

emf = -N * A * (dB/dt)

Where emf is the induced emf, N is the number of turns in the coil, A is the area of the coil, and (dB/dt) is the rate of change of magnetic field with time.

In this case, the initial area of the coil is given as 0.250 m², the number of turns is 50, and the strength of the uniform magnetic field is 1.50 T.

The coil is squished to have no area in 0.100 s, so the rate of change of area with time is 0.250 m² / 0.100 s = 2.50 m²/s.

Plugging these values into the formula, we get:

emf = -50 * 0.250 * 2.50

= -31.25 V

The negative sign indicates that the induced emf opposes the change in magnetic flux through the coil.

The direction of the induced emf is given by Lenz's Law, which states that the direction of the induced current generates a magnetic field that opposes the change in magnetic flux.

In this case, since the magnetic field is directed into the page, the induced current would generate a magnetic field directed out of the page, opposing the initial magnetic field.

User MamathaMacherla
by
8.6k points