A 2-meter long solenoid which may be considered to be infinitely long, is unifor

Question

A 2-meter long solenoid which may be considered to be infinitely long, is uniformly winded 6000 turns by a wire carrying current i1. The diameter of the solenoid is 20 cm.

a. a.     Inside the solenoid, another circular wire loop of diameter 15 cm and resistance 600 ?is oriented with its normal parallel to the solenoid axis. The cross section view is given in the figure. At the moment when solenoid current i1 is 4 A and increases at a rate of +10 A/s what is the magnitude of the induced current i2 in the circular wire loop? Show the direction of the induced current i2 if the solenoid current i1 runs clockwise.

b. a.     The above circular wire loop is replaced by a similarly oriented but larger wire loop of diameter 25cm and resistance 1000?. Determine the induced current in this wire loop which lies outside the solenoid when the solenoid current is 4 A and increases at a rate of +10 A/s.

A 2-meter long solenoid which may be considered to be infinitely long, is uniformly winded 6000 turns by a wire carrying current i1. The diameter of the solenoid is 20 cm. inside the solenoid, another circular wire loop of diameter 15 cm and resistance 600 ?is oriented with its normal parallel to the solenoid axis. The cross section view is given in the figure. At the moment when solenoid current i1 is 4 A and increases at a rate of +10 A/s what is the magnitude of the induced current i2 in the circular wire loop? Show the direction of the induced current i2 if the solenoid current i1 runs clockwise.

Explanation / Answer

B= (mu)nI

E=A(dB/dt)

as B is (mu)*nI

and mu= 4(pi)*10(-7)

n =number of turns per unit length= 6000/2

I= current

so dB/dt = (mu) n*(dI/dt) (as mu & n are constant)

now dB/dt= rate of change of current =10A/s (given)

and A = area of wire loop = (pi)r^2 r =15/2=7.5 cm

r= 0.075 m

so by putting these values in E=A(dB/dt)

E= [(pi)0.075*0.075]*[4(pi)*10^(-7)]*(6000/2)*10

=6.66*10^(-4)

E=IR

i2=6.66*10^(-4)/600

1.11*10^(-6) A

as current is clockwise and increasing so by right hand rule dB/dt will be in direction same as B (as shown in fig).

the induced emf will oppose it so B due to induced emf will be on opposit derection.

so current will be anticlockwise

2 no field outside

= B= 0 always

so

dB/dt=0

E=A(dB/dt) = 0

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