Question

An elastic circular conducting loop expands at a constant rate over time such that its radius is given by \(r(t)=r_{0}+v t,\) where \(r_{0}=0.100 \mathrm{~m}\) and \(v=0.0150 \mathrm{~m} / \mathrm{s}\). The loop has a constant resistance of \(R=12.0 \Omega\) and is placed in a uniform magnetic field of magnitude \(B_{0}=0.750 \mathrm{~T}\), perpendicular to the plane of the loop, as shown in the figure. Calculate the direction and the magnitude of the induced current, \(i\) at \(t=5.00 \mathrm{~s}\).

Short answer

Answer: At \(t=5.00 ~\text{s}\), the induced current in the loop is \(1.026 \times 10^{-3} ~\text{A}\) in the counterclockwise direction.

Step-by-step solution

Write down the given values

We are given the initial radius \(r_0=0.100 ~\text{m}\), the radial expansion velocity \(v=0.0150 ~\text{m/s}\), the resistance \(R=12.0 \Omega\), the magnetic field \(B_0=0.750 ~\text{T}\), and the time at which we want to calculate the induced current \(t=5.00 ~\text{s}\).

Calculate the radius at time \(t\)

Using the equation \(r(t)=r_0+vt\), we can find the radius at time \(t\): \(r(5.00~\text{s}) = 0.100~\text{m} + (0.0150 ~\text{m/s})(5.00~\text{s}) = 0.175~\text{m}\)

Calculate the magnetic flux through the loop

The magnetic flux \(\Phi_B\) through the circular loop is given by \(\Phi_B = B_0 A\), where \(A\) is the area of the loop. For a circle of radius \(r\), the area is given by \(A = \pi r^2\). So at \(t = 5.00 ~\text{s}\), the flux is, \(\Phi_B(5.00~\text{s}) = B_0 \pi r^2(5.00~\text{s})=(0.750~\text{T})\pi(0.175~\text{m})^2=0.028774~\text{T} \cdot \text{m}^2\)

Find the rate of change of magnetic flux

We first differentiate the magnetic flux \(\Phi_B\) with respect to time: \(\frac{d\Phi_B}{dt} = B_0 \frac{d(\pi r^2)}{dt} = 2 \pi r B_0 \frac{dr}{dt}\) We know the rate of change of radius, \(\frac{dr}{dt}\), is the constant \(v=0.0150 ~\text{m/s}\). So, the rate of change of magnetic flux at \(t=5.00 ~\text{s}\) is: \(\frac{d\Phi_B}{dt}(5.00 ~\text{s}) = 2\pi(0.175~\text{m})(0.750~\text{T})(0.0150 ~\text{m/s})=0.012319~\text{T} \cdot \text{m}^2/\text{s}\)

Calculate the induced emf

Using Faraday's law of electromagnetic induction, the induced emf \(\epsilon\) is given by the negative rate of change of magnetic flux: \(\epsilon = -\frac{d\Phi_B}{dt}\). Therefore, at \(t=5.00 ~\text{s}\), the emf is: \(\epsilon(5.00~\text{s}) = -(-0.012319~\text{T} \cdot \text{m}^2/\text{s}) = 0.012319~\text{V}\)

Calculate the induced current

Using Ohm's law, the current \(i\) is given by \(i = \frac{\epsilon}{R}\). So, at \(t=5.00 ~\text{s}\), the current is: \(i(5.00~\text{s}) = \frac{0.012319~\text{V}}{12.0 \Omega} = 1.026 \times 10^{-3} ~\text{A}\)

Determine the direction of the induced current

According to Lenz's Law, the induced current will oppose the change in magnetic flux. In this case, the change in magnetic flux is increasing due to the increasing area of the loop. Therefore, the induced current will create a magnetic field opposing the original magnetic field. Given the original magnetic field is pointing into the plane of the loop, the induced current will create a field pointing out of the plane of the loop, making the induced current flow counterclockwise.

Final answer

At \(t=5.00 ~\text{s}\), the induced current in the loop is \(1.026 \times 10^{-3} ~\text{A}\) in the counterclockwise direction.