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:

  • $ \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})} $



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