A metal bar of length L = 4.6 m slides along two horizontal metal rails. A magnetic field of magnitude B = 1.6 T is directed vertically. (a) If the bar is moving at speed v = 0.46 m/s, what is the emf induced across the ends of the bar? (in V) (b) Which end of the bar is at the higher potential? The end coming out of the page or the end going into the page?

Answers

Answer 1

A metal bar of length L slides along two horizontal metal rails with a magnetic field of magnitude B directed vertically. At a speed of v, the emf induced across the ends of the bar can be calculated and the end with the higher potential can be determined.

(a) The emf induced across the ends of the bar is given by the equation:

emf = BLv

where B is the magnetic field strength, L is the length of the bar, and v is the velocity of the bar. Substituting the given values, we get:

emf = (1.6 T)(4.6 m)(0.46 m/s) = 3.2 V

Therefore, the emf induced across the ends of the bar is 3.2 V.

(b) The direction of the emf induced across the bar is given by Lenz's law, which states that the direction of the induced emf is such that it opposes the change in magnetic flux that produced it. In this case, the magnetic field is directed vertically, so the magnetic flux through the bar is changing as it moves horizontally along the rails. By Fleming's right-hand rule, we can determine that the end of the bar going into the page is at the higher potential, and the end coming out of the page is at the lower potential.

To know more about the emf induced refer here :

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