Question

ring of resistance 10 ohm radius 10 cm and 100 turns is rotated at a rate of 100 revolutions per second about a fixed axis which is perpendicular to a uniform magnetic field of induction 10mt the amplitude of current in the loop will be nearly

Answer 1

The amplitude of the current is 9.86mA

When the coil  of N turns, area A is rotated in a magnetic field of strength B with an angular velocity ω, then the induced current is given by the expression,

$I=\frac{NBA\omega }{R} sin \omega t$

The amplitude of the current is given by,

$I=\frac{NBA\omega }{R}$

Since $\omega =2\pi f$, where f is the frequency of rotation of the coil, and $A=\pi r^2$, where r is the radius of the coil, then,

$I=\frac{NBA\omega }{R}\\ I=\frac{NB(\pi r^2)2\pi f }{R}$

Substituting the given values,

$I=\frac{NB(\pi r^2)2\pi f }{R}\\ =\frac{(100)(10*10^-^3T)(3.14)(10*10^-^2m)^2(2)(3.14)(100 rev/s)}{10 \Omega} \\ =9.86*10^-^3A\\ =9.86mA$

The amplitude of the current is 9.86mA

Answer 2

Explanation:

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