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The inductance of a closely wound coil is such that an emf of 3.00mVis induced when the current changes at the rate of 5.00A/s. A steady current of 8.00Aproduces a magnetic flux of 40.0 mWbthrough each turn. (a) Calculate the inductance of the coil. (b) How many turns does the coil have?

Short Answer

Expert verified
  1. The inductance of the coil isL=6×10-4H
  2. The number of turns of the coil isN=120

Step by step solution

01

Given

ε=3.00mV=0.003Vdidt=5.00A/si=8.00AÏ•B=40.0μ°Â²ú

02

Understanding the concept

By using the equation 30-35 we can find the inductance of the coil.If we pass a current iin the turns of the solenoid we can consider it as an inductor, the current produces a magnetic fluxϕBthrough the central region of the inductor. The inductance of the inductor is then defined interms of that flux as by equation 30-28.

Formula:

ε=-LdidtL=NϕBi

03

Calculate the inductance of the coil.

By using the equation 30-35 we can find the inductance

ε=-Ldidt

Rearrange as

L=εdidtL=0.0035.00L=6×10-4H

04

(b) Calculate the number of turns of the coil.

According to equation 30-28 the inductance can be defined as

L=NϕBi

We have to find the number of turns so the formula rearrange as

N=LiϕB

By substituting the value we get,

N=6×10-4×840×10-6N=120

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