Question

A small loop of area A = 5.5 mm2 is inside a long solenoid that has n = 919 turns/cm and carries a sinusoidally varying current i of amplitude 3.08 A and angular frequency 226 rad/s. The central axis of the loop and solenoid coincide. What is the amplitude of the emf induced in the loop?

Answer 1

Answer:
`410.90* 10^(-6) V`

Step-by-step explanation:

Given

`A=5.5 mm^2`

`n=919 turns/cm\approx 85400 turns/m`

`\omega =226 rad/s`

According to the Faraday law of induction, induced emf is given by

`E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}`

Magnetic field
`\phi _B=B\cdot A`

`B=\mu _0ni=\mu _0ni_osin\left ( \omega t\right )`

`B=4\pi * 10^(-7)* 85400* 3.08* \sin (226t)`

`B=0.3305\sin (226t)`

`\phi _(B)=0.3305* \sin (226t)* 5.5* 10^(-6)`

`\phi _(B)=1.818* 10^(-6)\sin (226t)`

`E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}`

`E=-1.818* 10^(-6)* 226\cos (226t)`

`E=-410.90* 10^(-6)\cos (226t)`

Amplitude of EMF induced
`=410.90* 10^(-6) V`