Question

The component of the external magnetic field along the central axis of a 50 -turn coil of radius \(5.0 \mathrm{cm}\) increases from 0 to 1.8 T in 3.6 s. (a) If the resistance of the coil is \(2.8 \Omega,\) what is the magnitude of the induced current in the coil? (b) What is the direction of the current if the axial component of the field points away from the viewer?

Short answer

Question: A circular coil with 50 turns, a radius of 5.0 cm, and a resistance of 2.8 Ω is subject to a magnetic field. Initially, the axial component of the magnetic field is zero, and it increases to 1.8 T within 3.6 seconds. Calculate the magnitude and direction of the induced current in the coil. Answer: The magnitude of the induced current is \(\frac{50\left(\frac{0.045\pi}{3.6}\right)}{2.8}\,\mathrm{A}\) and its direction is clockwise when viewed from above, as per Lenz's Law.

Step-by-step solution

Calculate the initial and final magnetic fluxes in the coil

Given the coil's radius \(r = 5.0 \mathrm{cm}\) (or 0.05 m) and the number of turns \(N = 50\), we will first find the area of the coil: \(A = \pi r^2\). The magnetic flux is given by \(\Phi = B\cdot A\), where B is the magnetic field component along the coil axis. Initial magnetic field \(B_{i} = 0\,\text{T}\) Final magnetic field \(B_{f} = 1.8\,\text{T}\) Calculating the initial and final magnetic fluxes, we have: \(\Phi_{i} = B_{i}\cdot A = 0 \cdot \pi (0.05)^2 = 0\,\text{Wb}\) \(\Phi_{f} = B_{f}\cdot A = 1.8 \cdot \pi (0.05)^2 = 0.045\pi\,\text{Wb}\)

Calculate the change in magnetic flux

To find the change in magnetic flux, subtract the initial magnetic flux from the final magnetic flux: \(\Delta\Phi = \Phi_{f} - \Phi_{i} = 0.045\pi\,\text{Wb}\)

Apply Faraday's Law

Faraday's Law states that the induced emf (electromotive force) is equal to the negative rate of change of magnetic flux through the coil: \(E = -N\frac{d\Phi}{dt}\) Given the change in magnetic flux, we have \(\Delta\Phi = 0.045\pi \,\mathrm{Wb}\). This change in flux occurs in 3.6 seconds, so the rate of change of magnetic flux is \(\frac{d\Phi}{dt} = \frac{\Delta\Phi}{\Delta t} = \frac{0.045\pi}{3.6}\,\mathrm{Wb/s}\), Now, we can find the emf: \(E = -N\frac{d\Phi}{dt} = -50\left(\frac{0.045\pi}{3.6}\right)\,\mathrm{V}\) The negative sign indicates the induced emf opposes the change in magnetic flux. However, we only need the magnitude of the emf: \(|E| = 50\left(\frac{0.045\pi}{3.6}\right)\,\mathrm{V}\)

Use Ohm's Law to find the induced current's magnitude

Ohm's Law states that the current \(I\) is equal to the emf \(E\) divided by the coil's resistance \(R\): \(I = \frac{E}{R}\) Given that the coil's resistance is \(R = 2.8\,\Omega\), the induced current's magnitude is: \(|I| = \frac{50\left(\frac{0.045\pi}{3.6}\right)}{2.8}\,\mathrm{A}\)

Determine the direction of the induced current using Lenz's Law

Lenz's Law states that the direction of the induced current will be such that it opposes the change in the magnetic flux that produced it. In our case, if the axial component of the field points away from the viewer, and the magnetic field component is increasing over time, then the induced current will create a field pointing toward the viewer to oppose this change. This can be achieved by a current flowing clockwise when viewed from above, following the right-hand rule (curl the fingers of the right hand in the direction of the induced current, and the thumb will point in the direction of the induced field).