A conducting bar slides without friction on two parallel horizontal rails that a

Question

A conducting bar slides without friction on two parallel horizontal rails that are 1.60 m apart and connected by a wire containing a 4.00 Ohm resistor at the left-hand end. At the instant seen in the diagram, the resistance of the bar and the rails is equal to 2.00 Ohm, and the current running through the circuit is 0.124 A. A uniform magnetic field is perpendicular to the plane of the rails. The bar is moving at a constant speed of 0.50 m/s. What is the magnitude of the magnetic field? a) 0.10 T b) 0.25 T c) 0.38 T d) 0.54 T e) 0.93 T

Explanation / Answer

Given

length of the bar l = 1.6 m

total resistance of the bar is 2 ohm , and resistance of the wire is 4 ohm

net resistance in the circuit is R = 2+4 = 6 ohm

current i = 0.124 A

bar moving with constant speed v = 0.5 m/s

magnitude of the magnetic field B = ?

this is like motional emf e = B*v*l ==> B = e/v*l

her we have to calculate the emf , from the given data e = i*R = 0.124*6 = 0.744 V

the magnetic field is B = e/v*l = 0.744/(0.5*1.6) = 0.93 T

the answer is option e) 0.93 T

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