Shows a conducting bar of length l = 0.7 m moving to the right on two frictionle

Question

Shows a conducting bar of length l = 0.7 m moving to the right on two frictionless rails. A uniform magnetic field directed into the page has magnitude B = 0.6 T. The resistor lias resistance R = 8.0 Ohm. The conducting bar is moving to the right at a constant velocity v = 10 m/s. Calculate: The induced (motional) emf epsilon in the rectangular loop formed by the rails, the conducting bar, and the resistor. The induced current I produced by the emf epsilon in the loop. The power delivered to the resistor. What is the direction of the current found in part b? Since the induced current I in the bar is in the magnetic field B. the bar is subject to a magnetic force to the left. In order for the bar to move to the right at a constant velocity, an external force must be applied to the right to balance this left-directed magnetic force. Calculate the magnitude of this magnetic force that the conducting bar experiences.

Explanation / Answer

Given that

Length of the bar (L) =0.7m

Magnitude of the uniform magnetic field is (B) =0.6T

The resistance of the resistor (R) =8ohm

Bar moving with a velocity (v) =10m/s

a)

The induced emf in the rectangular bar is (e) =BLv=(0.6T)(0.7m)(10m/s) =4.2V

b)

The induced current produced in the loop is given by ohm's law

e =IR

I =e/R =4.2V/8ohm =0.525A

c)

The power delivered to the resistor is

P =e2/R or P =I2*R =(0.525A)2(80hm)=2.205W

e)

The magnetic force that the conducting bar experience is given by

F =BIL =(0.6T)(0.525A)(0.7) = 0.2205N

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