Question

A solenoid made from 600 turns of wire on a cylindrical form 2cm in diameter and 10cm long has a current of 3 amps on it. Find the strength of the magnetic field on the axis of the coil. If the current in the solenoid is interrupted and falls to zero in .001 seconds, find the voltage induced across the coil

Answer 1

0.0226 T 4.25784 volt

Step-by-step explanation:

Length of the solenoid = 10 cm =0.1 m

Diameter of the solenoid d=2 cm , so radius
`r=(d)/(2)=(2)/(1)=1cm=0.01m`

Number of turns N = 600, so number of turns per unit length
`n=(N)/(L)=(600)/(0.1)=6000`

Current = 3 A

Magnetic field due to solenoid
`B=\mu _0ni=4\pi * 10^(-7)* 6000* 3=0.02261T`

Area
`A=\pi r^2=3.14* 0.01^2=3.14* 10^(-4)m^2`

Emf induced is given by
`e=-NA(dB)/(dt)=-600* 3.14* 10^(-4)((0-0226))/(0.001)=4.25784V`

Answer 2

0.0226 T, 7.096 V

Step-by-step explanation:

N = 600

l = 10 cm = 0.1 m

i = 3 A

diameter = 2 cm

The magnetic field on the axis is given by

B = μo x n x i

where, n be the number of turns per unit length, i be the current and μo be the permeability of free space.

`n =(N)/(l)= (600)/(0.1)=6000`

`B=4* 3.14 *  10^(-7)* 6000* 3`

B = 0.0226 T

time taken to decrease the magnetic field to be zero is t = 0.001 second

Let e be the amount of voltage induced.

According to the Faraday's law of electromagnetic induction

e = rate of change of magnetic flux

`e=(d\phi )/(dt)=A(dB)/(dt)`

where, A be the area of coil

`A = 3.14 * r^(2)= 3.14 * 0.01* 0.01=3.14* 10^(-4)m^(2)`

So, induced voltage

`e=(3.14* 10^(-4)* 0.0226 )/(0.001)=7.096* 10^(-3)V\\`

e = 7.096 mV