Question

The magnetic field at the centre of a toroid is 2.2-mT. If the toroid carries a current of 9.6 A and has 6.000 turns, what is the radius of the toroid?

Answer 1

The radius of the toroid is
`5.23*10^(-3)\ m`.

Step-by-step explanation:

Given that,

Magnetic field B = 2.2 mT

Current =9.6 A

Number of turns = 6.000

We need to calculate the radius

Using formula of magnetic force

`B=(\mu_(0)NI)/(2\pi r)`

`r=(\mu_(0)NI)/(2\pi B)`

Put the value int the formula

`r=(4\pi*10^(-7)*9.6*6.000)/(2\pi*2.2*10^(-3))`

`r=5.23*10^(-3)\ m`

Hence, The radius of the toroid is
`5.23*10^(-3)\ m`.

Answer 2

Radius, r = 0.00523 meters

Step-by-step explanation:

It is given that,

Magnetic field,
`B=2\ mT=2.2* 10^(-3)\ T`

Current in the toroid, I = 9.6 A

Number of turns, N = 6

We need to find the radius of the toroid. The magnetic field at the center of the toroid is given by :

`B=(\mu_oNI)/(2\pi r)`

`r=(\mu_oNI)/(2\pi B)`

`r=(4\pi * 10^(-7)* 6* 9.6)/(2.2\pi * 2* 10^(-3))`

r = 0.00523 m

or

`r=5.23* 10^(-3)\ m`

So, the radius of the toroid is 0.00523 meters. Hence, this is the required solution.