Question

A rectangular wire loop (dimensions of \(h=15.0 \mathrm{~cm}\) and \(w=8.00 \mathrm{~cm}\) ) with resistance \(R=5.00 \Omega\) is mounted on a door, as shown in the figure. The Earth's magnetic field, \(B_{\mathrm{E}}=2.60 \cdot 10^{-5} \mathrm{~T}\), is uniform and perpendicular to the surface of the closed door (the surface is in the \(x z\) -plane). At time \(t=0\), the door is opened (right edge moves toward the \(y\) -axis) at a constant rate, with an opening angle of \(\theta(t)=\omega t,\) where \(\omega=3.50 \mathrm{rad} / \mathrm{s}\). Calculate the direction and the magnitude of the current induced in the loop, \(i(t=0.200 \mathrm{~s})\).

Short answer

At t = 0.200 seconds, the induced current in the loop is given by: \(i(0.200 \mathrm{~s}) = \frac{0.0120 \mathrm{m}^2 \cdot 2.60 \times 10^{-5} \mathrm{T} \cdot (3.50 \mathrm{rad/s} \cdot \sin{(3.50 \mathrm{rad/s} \cdot 0.200\mathrm{s})})}{5.00 \Omega}\) Calculate the induced current, and using Lenz's law, the direction of the induced current is found to be clockwise when looking at the door from an opposite perspective to the door opening direction.

Step-by-step solution

Determine Magnetic Field Component Through the Loop

Initially, the Earth's magnetic field is perpendicular to the closed door (the surface is in the xz-plane). To find the component of the magnetic field that goes through the loop, we need to consider the change in the angle of the door with respect to time. At time \(t\), the opening angle of the door is given by: \(\theta(t) = \omega t\) At \(t=0.200 \mathrm{~s}\), the opening angle is: \(\theta(0.200 \mathrm{~s}) = \omega \cdot 0.200 = 3.50 \mathrm{rad/s} \cdot 0.200 \mathrm{s} = 0.700 \mathrm{rad}\) Since the Earth's magnetic field is constant and uniform, the component of the magnetic field through the loop is given by: \(B_\text{through} = B_\text{E} \cos(\theta)\) For \(t = 0.200 \mathrm{~s}\): \(B_\text{through}(0.200 \mathrm{~s}) = 2.60 \times 10^{-5} \mathrm{T} \cos(0.700 \mathrm{rad})\)

Calculate the Magnetic Flux Through the Loop

At any time t, the magnetic flux through the loop Φ(t) is given by the product of the area of the loop A, the magnetic field component through the loop \(B_\text{through}(t)\), and the cosine of the angle between the magnetic field and the normal to the loop: \(\Phi(t) = A \cdot B_\text{through}(t)\) Since the loop is rectangular, its area is given by: \(A = h \cdot w = 0.150 \mathrm{m} \times 0.080 \mathrm{m} = 0.0120 \mathrm{m}^2\) At \(t=0.200 \mathrm{~s}\), the magnetic flux is: \(\Phi(0.200 \mathrm{~s}) = 0.0120 \mathrm{m}^2 \cdot 2.60 \times 10^{-5} \mathrm{T} \cos(0.700 \mathrm{rad})\)

Calculate the Induced Electromotive Force (emf)

According to Faraday's law of electromagnetic induction, the induced emf ε in a loop is the negative time rate of change of the magnetic flux: \(\epsilon = -\frac{d\Phi}{dt}\) Taking the derivative of the flux with respect to time gives: \(\epsilon = -A \cdot \frac{d B_{\text{through}}(t)}{dt}\) Since only the opening angle θ(t) depends on time, this becomes \(\epsilon = -A \cdot B_E \cdot \left(-\omega \sin(\omega t)\right)\) At \(t = 0.200 \mathrm{~s}\), the induced emf is: \(\epsilon(0.200 \mathrm{~s}) = 0.0120 \mathrm{m}^2 \cdot 2.60 \times 10^{-5} \mathrm{T} \cdot (3.50 \mathrm{rad/s} \cdot \sin{(3.50 \mathrm{rad/s} \cdot 0.200\mathrm{s})})\)

Calculate the Induced Current

Using Ohm's law, we can find the induced current in the loop: \(i = \frac{\epsilon}{R}\) At \(t = 0.200 \mathrm{~s}\), the induced current is: \(i(0.200 \mathrm{~s}) = \frac{0.0120 \mathrm{m}^2 \cdot 2.60 \times 10^{-5} \mathrm{T} \cdot (3.50 \mathrm{rad/s} \cdot \sin{(3.50 \mathrm{rad/s} \cdot 0.200\mathrm{s})})}{5.00 \Omega}\) Finally, Lenz's law states that the direction of the induced current is such that it opposes the change in magnetic flux through the loop. Since the door is opening and the magnetic field is decreasing through the loop, the induced current will circulate clockwise from the perspective opposite to the door opening direction.