JEE Main & Advanced
Physics
Electro Magnetic Induction
Question Bank
Self Evaluation Test - Electromagnetic Induction
question_answer
A sliding wire of length 0.25 m and having a resistance of \[0.5\Omega \] moves along conducting guiding rails AB and CD with a uniform speed of 4 m/s. A magnetic field of 0.5 T exists normal to the plane of ABCD directed into the page. The guides are short -circuited with resistances of 4 and \[2\,\Omega \] as shown. The current through the sliding wire is:
A)0.27 A
B)0.37 A
C)1.0 A
D)0.72A
Correct Answer:
A
Solution :
[a] The induced emf across the sliding wire \[e=Bv\ell =0.5\times 4\times 0.25=0.5\,V\] The effective circuit is shown in figure. The equivalent resistance of the circuit \[r=\frac{4\times 2}{4+2}+0.5=1.83\Omega \] Now, \[i=\frac{V}{R}=\frac{0.5}{1.83}=0.27\,A\]