Question

A ring with an 18mm diameter falls off a scientist's finger into the solenoid in the lab. The solenoid is 25 cm long, 5.0 cm in diameter and has 1500 turns. When turned on, the current in the solenoid is increases linearly to 20 A in 1 second. What is the induced emf in the ring

Answer 1

The value is
`\epsilon = 3.84 *10^(-5) \ V`

Step-by-step explanation:

From the question we are told that

The diameter of the ring is
`d = 18 \ mm = 0.018 \ m`

The length of the solenoid is
`l = 25 \ cm = 0.25 \ m`

The diameter of the solenoid is
`D = 5.0 \ cm = 0.05 \ m`

The number of turns is N = 1500

The change in current in the solenoid is
`\Delta I = 20 \ A`

The time taken is
`\Delta t = 1 \ s`

Generally the radius of the ring is

`r = (d)/(2)`

=>
`r = (0.018 )/(2)`

=>
`r = 0.009 \ m`

Generally the area of the ring is mathematically represented as

`A = \pi r^2`

=>
`A = 3.142 * 0.009^2`

=>
`A = 2.545 *10^(-4)\ m^2`

Generally the induced emf is mathematically represented as

`\epsilon = A * (dB)/(dt)`

Here

`(dB )/(dt) = \mu_o * (N)/(l) *( \Delta I )/(\Delta t)`

Here
`\mu_o` is the permeability of free space with value

`\mu_o = 4\pi *10^(-7) \ N/A^2`

So

`(dB )/(dt) = 4\pi * 10^(-7) * (1500)/(0.25) *(20 )/(1)`

=>
`(dB )/(dt) = 0.150816\ T/s`

So

`\epsilon = 0.150816 * 2.545 *10^(-4)`

=>
`\epsilon = 3.84 *10^(-5) \ V`