Question

An 1300-turn coil of wire that is 2.2 cmcm in diameter is in a magnetic field that drops from 0.14 TT to 0 TT in 9.0 msms . The axis of the coil is parallel to the field.

What is the emf of the coil? (in V)

Answer 1

The induced emf is
`\epsilon =7.68 \ V`

Step-by-step explanation:

From the question we are told that

The number of turns is
`N = 1300 \ turns`

The diameter is
`d = 2.2 \ cm = 2.2*10^(-2)`

The initial magnetic field is
`B_i = 0.14 \ T`

The final magnetic field is
`B_f = 0 \ T`

The time taken is
`dt = 9.0ms = 9.0*10^(-3) \ s`

The radius is mathematically evaluated as

`r = (d)/(2 )`

substituting values

`r = (2.2 *10^(-2))/(2 )`

`r = 1.1*10^(-2) \ m`

The induced emf is mathematically represented as

`\epsilon =- N * (d\phi )/(dt )`

Where
`d\phi` is the change in magnetic field which is mathematically represented as

`d\phi = dB * A * cos\theta`

=>
`d\phi = [B_f - B_i ] * A * cos\theta`

Here
`\theta = 0` given that the axis of the coil is parallel to the field

Also A is the cross-sectional area which is mathematically represented as

`A = \pi r^2`

substituting values

`A = 3.142 * [1.1*10^(-2)]^2`

`A = 3.8 *10^{-4] \ m^2`

So

`d\phi = [0 - 0.14 ] * 3.8*10^(-4)`

`d\phi = -5.32*10^(-5) \ weber`

So

`\epsilon =- 1300 * (-5.32*10^(-5) )/( 9.0*10^(-3) )`

`\epsilon =7.68 \ V`