1. A conducting rod is pulled horizontally along a set of rails separated by d =

Question

1. A conducting rod is pulled horizontally along a set of rails separated by d= 0.220 m. A uniform magnetic field B= 0.500 T is directed into the page. There is no friction between the rod and the rails, and the rod moves with constant velocity v= 5.50 m/s. Using Faraday's Law, calculate the induced emf magnitude around the loop in the figure that is caused by the changing flux.

2. The emf around the loop causes a current to flow. If you are pulling the bar with a force of 3.200 N, how large is that current?

3. From your previous results, what must be the electrical resistance of the loop? (Assume that the resistance of the rails is negligible compared to the resistance of the rod, so the resistance of the loop is constant.)

A conducting rod is pulled horizontally along a set of rails separated by d= 0.220 m. A uniform magnetic field B= 0.500 T is directed into the page. There is no friction between the rod and the rails, and the rod moves with constant velocity v= 5.50 m/s. Using Faraday's Law, calculate the induced emf magnitude around the loop in the figure that is caused by the changing flux. The emf around the loop causes a current to flow. If you are pulling the bar with a force of 3.200 N, how large is that current? From your previous results, what must be the electrical resistance of the loop? (Assume that the resistance of the rails is negligible compared to the resistance of the rod, so the resistance of the loop is constant.)

Explanation / Answer

First we can establish the direction of the induced emf. Asthe rail moves to the right, the flux through the loop (into thepage or screen) increases because the loop gets bigger. The inducedemf will try to oppose this, creating a mag field that points outof the page. Using the right hand rule, with thumb pointing out ofthe page, we see the inducedemf is counterclockwise (direction of the curl of yourfingers). Now... for the magnitude induced emf = d (BA) / dt = B dA/dt = B d(width * d) /dt = = B d dw/dt = B dv because v, the speed, is therate at which the width is changing. So induced emf = B d v = 0.500 * 0.220 * 5.50 = 0.605 volts


B) current I=F/(Bd)= 29.09 A

R=V/I=0.605/29.09=20.8 m ohms
First we can establish the direction of the induced emf. Asthe rail moves to the right, the flux through the loop (into thepage or screen) increases because the loop gets bigger. The inducedemf will try to oppose this, creating a mag field that points outof the page. Using the right hand rule, with thumb pointing out ofthe page, we see the inducedemf is counterclockwise (direction of the curl of yourfingers). Now... for the magnitude induced emf = d (BA) / dt = B dA/dt = B d(width * d) /dt = = B d dw/dt = B dv because v, the speed, is therate at which the width is changing. So induced emf = B d v = 0.500 * 0.220 * 5.50 = 0.605 volts


B) current I=F/(Bd)= 29.09 A

R=V/I=0.605/29.09=20.8 m ohms

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