Electro Magnetic Induction And Alternating Currents Question 516
Question: There are three wire MO, NO and PQ, wires MO and NO are fixed and perpendicular to each other. Wire PQ moves with a constant velocity v as and resistance per unit length of each wire is $ \lambda $ and magnetic field exists perpendicular and inside the paper then. Which of the following is wrong?
Options:
A) current in loop is anticlockwise
B) magnitude of current in the loop is $ \frac{Bv}{\lambda (\sqrt{2}+1)} $
C) current in the loop is independent of time.
D) magnitude of current decreases as time increases.
Show Answer
Answer:
Correct Answer: D
Solution:
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$ \phi =BA=\frac{B}{2}\frac{x}{\sqrt{2}}\frac{x}{\sqrt{2}} $
$ \varepsilon =\frac{d\phi }{dt}=\frac{2x}{4}\frac{dx}{dt}B=\frac{x}{2}Bv $
$ \varepsilon =\frac{d\phi }{dt}=\frac{2x}{4}\frac{dx}{dt}B=\frac{x}{2}B\frac{d(2y)}{dt} $
$ i=\frac{xBv}{2\lambda (x+\sqrt{2}x)}=\frac{Bv}{\lambda (1+\sqrt{2})} $
