Electro Magnetic Induction And Alternating Currents Question 516
There are three wires MO, NO and PQ, wires MO and NO are fixed and perpendicular to each other. Wire PQ moves with a constant velocity v. The resistance per unit length of each wire is $ \lambda $. A 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}x\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})} $
BETA
