Question 18.7: Lighting a bulb A 10-Ω lightbulb is connected between the en......
Lighting a bulb
A 10-Ω lightbulb is connected between the ends of two parallel conducting rails that are separated by 1.2 m, as shown in the figure below. A metal rod is pulled along the rails so that it moves to the right at a constant speed of 6.0 m/s. The two rails, the lightbulb, its connecting wires, and the rod make a complete rectangular loop circuit. A uniform 0.20-T magnitude B_{\mathrm{ex}} field points downward, perpendicular to the loop’s area. Determine the direction of the induced current in the loop, the magnitude of the induced emf, the magnitude of the current in the lightbulb, and the power output of the lightbulb.
Sketch and translate We sketch the situation as shown, top right. Choose the normal vector for the loop’s area to point upward. The loop’s area increases as the rod moves away from the bulb. Because of this, the magnitude of the magnetic flux through the loop’s area is increasing as the rod moves to the right.
Simplify and diagram Assume that the rails, rod, and connecting wires have zero resistance. The induced field \vec{B}_{\text {in }} due to the loop’s induced current should point upward, resisting the change in the downward increasing magnetic flux through the loop’s area. Using the righthand rule for the \vec{B}_{\text {in }} field, we find that the direction of the induced current is counterclockwise.
Represent mathematically To find the magnitude of the induced emf, use Faraday’s law:
\varepsilon_{\text {in }}=N\left|\frac{\Phi_{\mathrm{f}}-\Phi_{\mathrm{i}}}{t_{\mathrm{f}}-t_{\mathrm{i}}}\right|
The angle between the loop’s normal line and the B_{\mathrm{ex}} field is 180°. The magnitude of the B_{\mathrm{ex}} field is constant. Therefore:
\varepsilon_{\text {in }}=(1)\left|\frac{B_{\mathrm{ex}} A_{\mathrm{f}} \cos \left(180^{\circ}\right)-B_{\mathrm{ex}} A_{\mathrm{i}} \cos \left(180^{\circ}\right)}{t_{\mathrm{f}}-t_{\mathrm{i}}}\right|
=B_{\mathrm{ex}}\left|\frac{A_{\mathrm{f}}-A_{\mathrm{i}}}{t_{\mathrm{f}}-t_{\mathrm{i}}}\right|
The area A of the loop at a particular clock reading equals the length L of the sliding rod times the x-coordinate of the sliding rod (the origin of the x-axis is placed at the bulb.) Therefore:
\varepsilon_{\text {in }}=B_{\text {ex }}\left|\frac{L x_{\mathrm{f}}-L x_{\mathrm{i}}}{t_{\mathrm{f}}-t_{\mathrm{i}}}\right|=B_{\mathrm{ex}} L\left|\frac{x_{\mathrm{f}}-x_{\mathrm{i}}}{t_{\mathrm{f}}-t_{\mathrm{i}}}\right|=B_{\mathrm{ex}} L v
The quantity inside the absolute value is the x-component of the rod’s velocity, the absolute value of which is the rod’s speed v. Using Ohm’s law, the induced current depends on the induced emf and the bulb resistance:
I_{\text {in }}=\frac{\varepsilon_{\text {in }}}{R}
The power output of the lightbulb will be
P=I_{\mathrm{in}}^2 R
Solve and evaluate The magnitude of the induced emf around the loop is
\varepsilon_{\text {in }}=B_{\mathrm{ex}} L v=(0.20 \mathrm{~T})(1.2 \mathrm{~m})(6.0 \mathrm{~m} / \mathrm{s})=1.44 \mathrm{~V}
The current in the bulb is
I_{\text {in }}=\frac{1.44 \mathrm{~V}}{10 \Omega}=0.14 \mathrm{~A}
The lamp should glow, but just barely, since its power output is only
P=I_{\text {in }}{ }^2 R=(0.14 \mathrm{~A})^2(10 \Omega)=0.21 \mathrm{~W}
Try it yourself: Suppose the rod in the last example moves at the same speed but in the opposite direction so that the loop’s area decreases. Determine the magnitude of the induced emf, the magnitude of current in the bulb, and the direction of the current.
Answer: 1.4 V, 0.14 A, and clockwise.