What is the emf produced in a 1.5 meter wire moving at a speed of 6.2 meters/second perpendicular to a magnetic field of strength 3.96 × 10-3 newtons/amp·meter ?

Answer : The emf produced is, 0.0368 volt

Solution :

Formula used :

This equation is valid when B, l and v are mutually perpendicular to each other.

where,

B = magnetic field of strength =

l = length of wire = 1.5 m

v = velocity = 6.2 m/s

Now put all the given values in the above formula, we get the emf.

Therefore, the emf produced is, 0.0368 volt

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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