Question

Q6].  

A coil with 6 windings or turns, and an area of 0.0875m² rotates in a uniform magnetic field of magnitude 0.963T. The emf induced in the coil has the following form as a function of time :

ε(t)= ε₀ sin(ωt)

ε₀ = 11 Volts. Calculate ω in radians per second.

Q7]. The current in a long solenoid with 12 turns per meter of radius 0.7007m depends on time as follows:

I_s= 33.1(Ampere/second)*t

where t is time.

A circular loop of wire of radius 0.198m is concentric around the solenoid. An electrical current of 0.00397 Amps is induced in the loop. (Solenoid can be treated as infinitely long. Note the wire loop is outside the solenoid.) Find the resistance of the loop in Ohms.

Q7A].  In the previous situation, in which a current through a solenoid is increased and thereby a current is induced in a wire loop which is concentric with the solenoid. Is the induced current in the loop in the same direction as the current in the solenoid or in the opposite direction?

Answer 1

6)Peak value of emf E0=BAN

B=magnetic field=0.963T

A=area of coil=0.0875m2

N=number of turns=6

=angular velocity=?

E0=11Volt

E0=BAN

=E0/BAN=11/(.963x.0875x6)=21.757rad/sec

7)Induced emf in the loop(e)=d/dt

=magnetic flux=BA

B=magnetic field in solenoid=0nI

0=4x10^-7

n=number of turns per unit length of solenoid=12

I=Is=33.1t(given)

B=0nI=4x(10^-7)x12x33.1t=4988.83x(10^-7)t (10^-7 means -7 is power of 10))

A=area of loop=r^2=3.14x(.198)^2=3.14x.198x.198=0.123m2

=BA=4988.83x(10^-7)tx.123=648.548x(10^-7)t

Induced emf in the loop(e)=d/dt=d[648.548x(10^-7)t]/dt=648.548x(10^-7)dt/dt=6.485x(10^-5)volt   

Induced current(i)=0.00397A

i=e/R

R=resistance =e/i=6.485x(10^-5)/0.00397=0.0163

7A) When the current in solenoid increases, induced current opposes change in applied current (by Lenz's law).Hence induced current is in opposite diretion of applied current