Question

The current in a long solenoid of radius 5 cm and 17 turns/cm is varied with time at a rate of 5 A/s. A circular loop of wire of radius 7 cm and resistance 5 Ω surrounds the solenoid. Find the electrical current induced in the loop (in µA).

Answer 1

The current is
`I =0.2mA`

Step-by-step explanation:

From the question we are told that

The first radius is
`R_1 = 5cm = (5)/(100) = 0.05cm`

The number of turns is
`N = 17 \ turn/cm`

The current rate is
`(dI)/(dt) = 5 A/s`

The second radius is
`R_2 = 7cm = (7)/(100) = 0.07m`

The resistance is
`r = 5 \Omega`

Generally the magnetic flux induced in the solenoid is mathematically represented as

`\O = B A`

Where is the magnetic field mathematically represented as

`B = N \mu_o I`

Where
`\mu_o` is the permeability of free space with a value of
`\mu_o = 4\pi *10^(-7) N/A^2`

and A is the area mathematically represented as

`A = \pi (R_2 - R_1)^2`

So

`\O = N \mu I * \pi R^2`

Substituting values

`\O = 17 * 4\pi *10^(-7) * \pi (7-5)^2I`

`\O = 2.68*10^(-4)I`

The induced emf is mathematically represented as

`\epsilon =- |(d\O)/(dt)|`

`\epsilon = 2.68*10^(-4 ) (dI)/(dt)`

substituting values

`\epsilon =2.68 *10^(-4) * 5`

`=1.3 *10^(-3) V`

From Ohm law

`I = (\epsilon )/(r)`

Substituting values

`I = (1.3*0^(-3))/(5)`

`I =0.2mA`