Question

Question #4 A. [1 point] Explain briefly the factors affecting the magnitude of the induced emf in an electricity generating station. B. [4 Points] A rectangular coil of 10 turns with dimensions 10 ?? × 8 ?? sides is held perpendicular to the uniform magnetic field, with half the area of the loop in the field, as shown the following figure. The loop contains an ideal battery with emf E = 1.5 ?. The magnitude of the magnetic field varies with time according to ? = − 4?2 + 3? + 3.0 , with B in teslas and t in seconds. (a) Calculatethemagnitudeanddirectionoftheemfinducedaroundtheloopat?=15???. (b) Find the net current with appropriate direction in the loop at ? = 15 ??? if the loop has a resistance of 2 ohm

Answer 1

A)Induced emf in the gnerating station E=BANsint

where B=magnetic field

A=area of coil

N=number of turns of coil

=angular velocity of rotation

t=time

B)Induced emf in the loop e=d/dt

=magnetic flux=BAN

B=magnetic field=-4+3t+3

A=area of loop=0.1x.08/2 =4xm2 (length=10cm=.1m, breadth=8cm. Half of loop is inside field)

N=number of turns=10

=BAN=(-4+3t+3)4xx10=0.04x((-4+3t+3)

Induced emf in the loop e=d/dt=d0.04x((-4+3t+3))/dt

e=0.04x[(-4x2t)+3]=-0.32t+.12=induced emf in the loop

At t=15sec, e=-(0.32x15)+.12=-4.68volt

Direction of induced emf is given by Lenz's law e=-d/dt=-(-4.68)=+4.68volt. (the figure is not shown in the question. Hence we cannot say about direction of magnetic field, which depends on direction of emf

induced current

Now we can consider 2 emfs 15V battery and induced emf 4.68V

Total voltage(V)=15+4.68=19.68V

resistance of loop=R=2

current I=V/R=19.68/2=9.84A