Question

A conducting rod which is hinged at center of a conducting loop of radius 'a' which is fixed in a horizontal plane. Another end of rod is sliding on conducting loop as shown in figure. A resistor R is connected between center and boundary of loop. When rod is rotated with angular speedo in vertical uniform magnetic field (B) then current i at given instant when PCQ is right angle (Considered negligible resistance of rod and loop) ​

Answer 1

Answer:

The current i at given instant when PCQ is right angle is given as

$i = \frac{1}{2R}B\omega_o a^2$

Explanation:

As per Faraday's law we know that the rate of change in magnetic flux is induced EMF in the loop

so we will have

$EMF = \frac{d\phi}{dt}$

here we know that

$\phi = B.\frac{1}{2}r^2\theta$

now we have

$\frac{d\phi}{dt} = B.\frac{1}{2}r^2\frac{d\theta}{dt}$

here we know that

$\frac{d\theta}{dt} = \omega_o$

so we have

$EMF = \frac{1}{2}B\omega_o a^2$

now Total current in the loop is given by

$i = \frac{V}{R}$

$i = \frac{1}{2R}B\omega_o a^2$

#Learn

Topic : Motional EMF

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