Question

A circular coil of one turn of radius 5.0 cm is rotated about a diameter with a constant angular speed of 80 revolutions per minute. A uniform magnetic field B = 0.010 T exists in a direction perpendicular to the axis of rotation. Find (a) the maximum emf induced, (b) the average emf induced in the coil over a long period and (c) the average of the squares of emf induced over a long period.

Answer 1

Answer:

rotated about a diameter with a constant angular speed of 80 revolutions per minute. A uniform magnetic field B = 0.010 T exists in a direction perpendicular to the axis of rotation. Find (a) the maximum emf induced, (b) the this line i can't understand

Answer 2

(a) The maximum emf induced is $6.66 \times 10^{-4} V$

(b) The average emf induced in the coil over a long period is zero

(c) The average of the squares of emf induced over a long period is $2.2 \times 10^{-7} V^{2}$

Explanation:

Circular spiral radius, R = 5.0 cm

Angular velocity of moving coil, ω= 80 revolutions/minute

Magnetic field perpendicular to the rotation axis, B = 0.010 T

We know that

The emf induced into the coil

$e=\frac{d \phi}{d t}$

$e=\frac{d B \cdot A \cos \theta}{d t}$

$e=-B A \sin \theta \frac{d \theta}{d t}$

$e=-B A \omega \sin \theta$

( $\frac{d \theta}{d t}=\omega=$ Level of angle shift between arc vector and B)

(a) In case of  maximum electromotive force,  

$\sin \theta=1$

    $e=B A \omega$

    $e=0.010 \times 25 \times 10^{-4} \times 80 \times \frac{(2 \pi \times \pi)}{60}$

    $e=0.66 \times 10^{-3}=6.66 \times 10^{-4} V$

(b) Each instant the induced emf direction changes. So the maximum emf is zero.

(c) The emf generated in the coil is $e=-B A \omega \sin \theta=-B A \omega \sin \omega t$

The average of emf induced squares is given by

$e_{a v}^{2}=\frac{\int_{0}^{T} B^{2} A^{2} W^{2} \sin ^{2} w t d t}{\int_{0}^{T} d t}$

$e_{a v}^{2}=\frac{B^{2} A^{2} W^{2} \int_{0}^{T} \sin ^{2} w t d t}{\int_{0}^{T} d t}$

$e_{a v}^{2}=\frac{B^{2} A^{2} W^{2} \int_{0}^{T}(1-\cos 2 w t) d t}{2 T}$

$e_{a v}^{2}=\frac{B^{2} A^{2} W^{2}}{2 T}\left[T-\frac{\sin 2 w t}{2 w}\right]_{0}^{T}$

$e_{a v}^{2}=\frac{B^{2} A^{2} W^{2}}{2 T}\left[T-\frac{\sin 4 \pi-\sin 0}{2 w}\right]=\frac{B^{2} A^{2} W^{2}}{2 T}$

$e_{a v}^{2}=\frac{\left(6.66 \times 10^{-4}\right)^{2}}{2}=22.1778 \times 10^{-8} V^{2}$

$\left(B A \omega=6.66 \times 10^{-4} V\right)$

$e_{a v}^{2}=2.2 \times 10^{-7} V^{2}$