Question

A fixed coil of wire with 10 turns and an area of 0.055 m2 is placed in a perpendicular magnetic field. This field oscillates in direction and magnitude at a frequency of 10 Hz and has a maximum value of 0.12 T, (a) What is the average emf induced in the coil during the time it takes for the field to go from its maximum value in one direction to its maximum value in the other direction? (b) Repeat part (a) for a time interval of one complete cycle. (c) At what time(s) during a complete magnetic field cycle would you expect the induced emf to have its maximum magnitude? What about its minimum value? Explain both answers

Answer 1

Answer:

Part a)

Average EMF for half cycle is

$E_{avg} = 2.64 V$

Part b)

For one complete cycle we will have

$E_{avg} = 0$

Part c)

Maximum induced EMF will be at

t = 0.025 s and 0.075 s

minimum induced EMF is at

t = 0.05s and 0.1 s

Explanation:

As we know that magnetic field is oscillating in direction as well as magnitude

so induced EMF is given as

$E = NBA\omega sin(\omega t)$

Part a)

For average value of EMF from positive maximum to negative maximum which is equal to half cycle

so we have

$E_{avg} = NBA\omega \frac{2}{T}\int_0^{T/2} sin(\omega t) dt$

$E_{avg} = \frac{2NBA\omega}{\pi}$

$E_{avg} = \frac{2(10)(0.12)(0.055)(2\pi (10))}{\pi}$

$E_{avg} = 2.64 V$

Part b)

For one complete cycle we will have

$E_{avg} = NBA\omega \frac{1}{T}\int_0^T sin(\omega t) dt$

$E_{avg} = 0$

Part c)

Maximum induced EMF will be at

$t = \frac{T}{4} and \frac{3T}{4}$

here we know

$T = \frac{1}{f} = 0.1 s$

t = 0.025 s and 0.075 s

minimum induced EMF is at

$t = \frac{T}{2} and T$

so it is

t = 0.05s and 0.1 s