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As as viewed along the central axis of the cylinder, a uniform magnetic field B is confined to the inner cylinder of radius 0.050 m. This field is directed perpendicular into the plane of the page and is decreasing at a steady rate of 0.900 T/s. Determine the magnitude and direction of the current induced in the outer circular wire ring of radius 0.16 m and resistance 7.1 Ω that encircles this magnetic field region.

Group of answer choices

1.0 × 10^-3 A, counterclockwise

4.0 × 10^-2 A, clockwise

4.0 × 10^-2 A, counterclockwise

0.13 A, clockwise

1.0 × 10-3 A, clockwise

1 Answer

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Final answer:

Using Faraday's law and Ohm's law to calculate the induced emf and current and applying Lenz's law for direction, the magnitude of the induced current is approximately 4.0 × 10^-4 A, and its direction is counterclockwise.

Step-by-step explanation:

The question involves finding the magnitude and direction of the induced current in a circular wire ring due to a changing magnetic field, which is a concept from physics involving Faraday's and Lenz's laws. To find the induced electromagnetic force (emf), we use Faraday's law which states that the induced emf (ε) in a loop is equal to the negative rate of change of magnetic flux (Φ) through the loop. The rate of change of the magnetic field is given as 0.900 T/s.

The magnetic flux (Φ) through the outer wire loop is Φ = B ⋅ A, where B is the magnetic field and A is the area of the loop. Since the field is only present in the inner cylinder, the relevant area is that of the inner cylinder (A = π ⋅ (0.050 m)^2). The change in magnetic flux is given by dΦ/dt = π ⋅ (0.050 m)^2 ⋅ (0.900 T/s). Calculating the induced emf gives ε = -dΦ/dt.

To find the induced current (I), Ohm's law is used, which relates the emf to current and resistance (I = ε/R). After determining the current's magnitude, we use Lenz's law to find the direction of the induced current which opposes the change in magnetic flux. Since the magnetic field is decreasing, the induced current will create a field to oppose the decrease, leading to a counterclockwise current when viewed from above (direction into the page).

Plugging in the numbers: ε = -(π ⋅ (0.050 m)^2 ⋅ 0.900 T/s) = -0.0028 V, and I = ε/R = -0.0028 V / 7.1 Ω = -3.94 × 10^-4 A. The negative sign indicates the direction is opposite to the field's decrease, which is counterclockwise.

Therefore, the magnitude of the induced current is about 4.0 × 10^-4 A, and the direction is counterclockwise.

User Yordan Nikolov
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