A 50-turn circular coil (radius = 15 cm) with a total resistance of 4.0 Ω is placed in a uniform magnetic field directed perpendicularly to the plane of the coil. The magnitude of this field varies with time according to B = A sin (αt), where A = 80 mT and α = 50π rad/s. What is the magnitude of the current induced in the coil at t = 20 ms?

Enter Answer in (milli Amps)

Question

A 50-turn circular coil (radius = 15 cm) with a total resistance of 4.0 Ω is placed in a uniform magnetic field directed perpendicularly to the plane of the coil. The magnitude of this field varies with time according to B = A sin (αt), where A = 80 mT and α = 50π rad/s. What is the magnitude of the current induced in the coil at t = 20 ms?

Enter Answer in (milli Amps)

BlackTom AI · Accepted Answer

We start by restating the given quantities and identifying what needs to be calculated. There are N = 50 turns, coil radius r = 0.15 m, so the cross-sectional area is A_loop = π r^2 = π (0.15)^2 = π × 0.0225 ≈ 0.070685 m^2. The coil resistance is R = 4.0 Ω. The external magnetic field is B(t) = A sin(α t) with A = 0.080 T and α = 50π rad/s. The time in question is t = 0.02 s (20 ms).

Step 1: Evaluate the magnetic field and its time rate at t = 0.02 s. Compute the argument α t = (50π)(0.02) = π, so B(t) = 0.080 sin(π) = 0 T. Although the field is momentarily zero, the emf depends on the time rate of change of the flux, not the instantaneous flux value.

Step 2: Compute dB/dt at t = 0.02 s. Since B(t) = A sin(α t), dB/dt = A α cos(α t). With α t = π, cos(π) = -1, so dB/dt = (0.080)(50π)(-1) = -4π ≈ -12.566 T/s.

Step 3: Compute the induced electromotive force (emf) using Faraday’s law. The magnetic flux through one loop is Φ = B A_loop, so dΦ/dt = A_loop dB/dt. The total emf for N turns is ε = -N dΦ/dt = -N A_loop dB/dt. Substituting the values gives ε = -50 × (0.070685 m^2) × (-4π T/s) = 50 × 0.070685 × 4π V ≈ 44.4 V.

Step 4: Determine the induced current magnitude. Ohm’s law for the coil is I = |ε| / R = 44.4 V / 4.0 Ω ≈ 11.1 A. Converting to milliamps, I ≈ 11.1 A × 1000 = 11100 mA.

Final note on units: the problem asks for the current in milliAmps, and the calculation yields approximately 11100 mA at t = 20 ms. The fact that B(t) = 0 at this instant does not imply zero emf, because the emf depends on the time derivative of B, which is nonzero here due to the changing phase of the sinusoid.

PHYS-222-30000 Chapter 23 - Faraday's Law and Inductance Test

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