The drawing shows a copper wire (negligible resistance) bentinto a circular shape with a radius of 0.48 m. The radial sectionBC is fixed in place, while the copper bar ACsweeps around at an angular speed of 16 rad/s. The bar makeselectrical contact with the wire at all times. The wire and the barhave negligible resistance. A uniform magnetic field existseverywhere, is perpendicular to the plane of the circle, and has amagnitude of 3.4 x 10-3 T. Find the magnitude of thecurrent induced in the loop ABC.

Question

The drawing shows a copper wire (negligible resistance) bentinto a circular shape with a radius of 0.48 m. The radial sectionBC is fixed in place, while the copper bar ACsweeps around at an angular speed of 16 rad/s. The bar makeselectrical contact with the wire at all times. The wire and the barhave negligible resistance. A uniform magnetic field existseverywhere, is perpendicular to the plane of the circle, and has amagnitude of 3.4 x 10-3 T. Find the magnitude of thecurrent induced in the loop ABC.

resistance = 3 ohms

I am having trouble figuring out the algebra for the equation.

I have the emf = -N(ΔF/Δt) but I don't know how to relate the angular speed into the equation. I could find the current with the equation

I= emf*r but I don't know how to find the emf properly

Answer 1

To find the magnitude of the current induced in the loop ABC, you can use Faraday's Law of Electromagnetic Induction. The equation you mentioned, emf = -N(ΔF/Δt), is a good starting point.

Here's how you can relate the angular speed to the equation:

1. Start by finding the magnetic flux through the loop. The magnetic flux (Φ) is equal to the product of the magnetic field (B) and the area (A) enclosed by the loop. In this case, the loop is a circle, so the area is πr^2, where r is the radius of the circle.

Φ = B * A = B * π * r^2

2. Now, determine the rate of change of the magnetic flux with respect to time. Since the bar AC is rotating and making electrical contact with the wire, it is changing the area of the loop. The rate of change of the area (dA/dt) is equal to the product of the angular speed (ω) and the length of the bar (AC). In this case, the length of the bar is the same as the circumference of the loop, which is 2πr.

dA/dt = ω * 2πr

3. Substitute the values of Φ and dA/dt into the equation for emf:

emf = -N * (dΦ/dt) = -N * (d(B * A)/dt) = -N * (B * dA/dt) = -N * B * (ω * 2πr)

4. Finally, use Ohm's Law (V = IR) to relate the emf to the current (I). Since the wire and bar have negligible resistance, the emf across the loop will be equal to the potential difference (V) across the loop. The potential difference is equal to the current multiplied by the total resistance (R = 3 Ω), which in this case is just the resistance of the loop.

emf = V = I * R

You can solve for I by equating the two emf equations:

-I * N * B * (ω * 2πr) = I * R

Now you can substitute the given values (N, B, ω, r, and R) into the equation and solve for the magnitude of the current (I).