question_answer

A straight wire conductor of length\[l\]of 0.4 m is moving with a speed\[v\]of 7 m/s perpendicular to a magnetic field B of intensity\[0.9\text{ }Wb/{{m}^{2}}\]. The induced emf across the conductor is

A)  2.52V         

B)  25.2V

C)  5.26V          

D)  1.26V

Correct Answer: A

Solution :

 Let a rod of length I moves with velocity\[v,\]in a magnetic field B. Because of change in magnetic flux passing through the circuit, an emf is induced in the circuit, it is given by \[e=Bvl\] Given, \[B=0.9\text{ }Wb/{{m}^{2}},\text{ }v=7\text{ }m/s,\text{ }I=0.4\text{ }m\] \[\therefore \] \[e=0.9\times 7\times 0.4=2.52V\]