Physics: Electromagnetic Induction and Faraday's Law
Calculating induced emf and predicting induced current direction
Physics: Electromagnetic Induction and Faraday's Law
Calculating induced emf and predicting induced current direction
Physics - Grade 9-12
- 1
A circular wire loop has an area of 0.25 m^2 and is placed perpendicular to a uniform magnetic field of 0.80 T. What is the magnetic flux through the loop?
Use magnetic flux = BA cos(theta), where theta is the angle between the magnetic field and the loop's area vector.
The magnetic flux is 0.20 Wb because flux equals BA cos(theta), and the field is perpendicular to the loop so theta = 0 degrees. The calculation is (0.80 T)(0.25 m^2)(1) = 0.20 Wb. - 2
A 50-turn coil experiences a change in magnetic flux from 0.10 Wb to 0.40 Wb in 0.20 s. What is the magnitude of the average induced emf?
The magnitude of the average induced emf is 75 V. Using emf = N(Delta flux)/(Delta time), the result is 50(0.40 Wb - 0.10 Wb)/(0.20 s) = 75 V. - 3
State Faraday's law in words and explain what factor makes the induced emf larger.
Focus on the rate of change of magnetic flux and the number of turns in the coil.
Faraday's law states that a changing magnetic flux through a circuit induces an emf in the circuit. The induced emf becomes larger when the magnetic flux changes more quickly or when the coil has more turns. - 4
A bar magnet is pushed north-pole first toward a coil connected to a light bulb. Explain why the bulb lights while the magnet is moving but not when the magnet is held still.
The bulb lights while the magnet is moving because the magnetic flux through the coil is changing, which induces an emf and current. When the magnet is held still, the flux is constant, so no emf is induced and the bulb does not light. - 5
A square loop with side length 0.20 m is in a magnetic field that increases from 0.50 T to 1.10 T in 0.30 s. The field is perpendicular to the loop. What is the magnitude of the induced emf?
First find the area of the square loop, then use the change in magnetic flux divided by time.
The induced emf has magnitude 0.080 V. The area is (0.20 m)(0.20 m) = 0.040 m^2, so the flux change is (1.10 T - 0.50 T)(0.040 m^2) = 0.024 Wb. Dividing by 0.30 s gives 0.080 V. - 6
According to Lenz's law, what is the direction of the induced magnetic field if the magnetic field into the page through a loop is increasing?
The induced effect opposes the change, not necessarily the original field.
The induced magnetic field points out of the page. Lenz's law says the induced field opposes the increase in flux, so it must point opposite the increasing field into the page. - 7
For the situation in Problem 6, is the induced current clockwise or counterclockwise as viewed on the page?
The induced current is counterclockwise. By the right-hand rule, a counterclockwise current produces a magnetic field out of the page. - 8
A 200-turn coil has an induced emf of 12 V when the magnetic flux through each turn changes uniformly for 0.50 s. What is the magnitude of the flux change per turn?
Rearrange emf = N(Delta flux)/(Delta time) to solve for Delta flux.
The flux change per turn is 0.030 Wb. Rearranging Faraday's law gives Delta flux = emf times Delta time divided by N, so Delta flux = (12 V)(0.50 s)/200 = 0.030 Wb. - 9
A straight conducting rod of length 0.40 m moves at 6.0 m/s perpendicular to a 0.25 T magnetic field. What motional emf is induced across the rod?
The motional emf is 0.60 V. For a rod moving perpendicular to the field, emf = BLv = (0.25 T)(0.40 m)(6.0 m/s) = 0.60 V. - 10
A coil is rotated in a uniform magnetic field. At one instant, the coil's area vector is parallel to the magnetic field. At another instant, the area vector is perpendicular to the magnetic field. At which instant is the magnetic flux greatest?
The angle in the flux equation is measured between the magnetic field and the area vector, not between the field and the surface of the coil.
The magnetic flux is greatest when the coil's area vector is parallel to the magnetic field. Since flux equals BA cos(theta), the maximum occurs when theta = 0 degrees and cos(theta) = 1. - 11
A single loop has an area of 0.060 m^2. The magnetic field through the loop changes from 0.20 T out of the page to 0.20 T into the page in 0.10 s. What is the magnitude of the average induced emf?
The magnitude of the average induced emf is 0.24 V. Treating out of the page as positive, the field changes from +0.20 T to -0.20 T, so the change in field is 0.40 T in magnitude. The flux change is (0.40 T)(0.060 m^2) = 0.024 Wb, and emf = 0.024 Wb/0.10 s = 0.24 V. - 12
A transformer has 400 turns on the primary coil and 100 turns on the secondary coil. If the primary voltage is 120 V AC, what is the secondary voltage, assuming an ideal transformer?
Use the transformer ratio Vs/Vp = Ns/Np.
The secondary voltage is 30 V AC. For an ideal transformer, Vs/Vp = Ns/Np, so Vs = 120 V times 100/400 = 30 V. - 13
Why does a transformer require alternating current instead of steady direct current for normal operation?
A transformer requires alternating current because a changing current in the primary coil produces a changing magnetic flux in the core. That changing flux induces an emf in the secondary coil, while steady direct current would produce nearly constant flux after it is established. - 14
A metal ring is falling vertically through a region where a horizontal magnetic field is present. As it enters the field region, the magnetic flux through the ring changes. Explain how electromagnetic induction affects the ring's motion.
Use Lenz's law to connect the induced current with a force that opposes the change causing it.
As the ring enters the magnetic field region, an induced current forms because the magnetic flux through the ring changes. By Lenz's law, the induced current creates a magnetic effect that opposes the change, which produces a force that resists the ring's motion and can slow it down. - 15
A coil has 80 turns and area 0.015 m^2 per turn. It is in a magnetic field perpendicular to the coil. The field decreases uniformly from 0.90 T to 0.10 T in 0.40 s. What is the magnitude of the induced emf?
Find the flux change for one turn first, then multiply by the number of turns.
The induced emf has magnitude 2.4 V. The change in flux per turn is (0.90 T - 0.10 T)(0.015 m^2) = 0.012 Wb. Multiplying by 80 turns and dividing by 0.40 s gives emf = 80(0.012 Wb)/(0.40 s) = 2.4 V.