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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?

Respuesta :

Answer:

Radius, r = 0.00523 meters

Explanation:

It is given that,

Magnetic field, [tex]B=2\ mT=2.2\times 10^{-3}\ T[/tex]

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 :                  

[tex]B=\dfrac{\mu_oNI}{2\pi r}[/tex]

[tex]r=\dfrac{\mu_oNI}{2\pi B}[/tex]  

[tex]r=\dfrac{4\pi \times 10^{-7}\times 6\times 9.6}{2.2\pi \times 2\times 10^{-3}}[/tex]  

r = 0.00523 m

or

[tex]r=5.23\times 10^{-3}\ m[/tex]

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

Answer:

The radius of the toroid is [tex]5.23\times10^{-3}\ m[/tex].

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

[tex]B=\dfrac{\mu_{0}NI}{2\pi r}[/tex]

[tex]r=\dfrac{\mu_{0}NI}{2\pi B}[/tex]

Put the value int the formula

[tex]r=\dfrac{4\pi\times10^{-7}\times9.6\times6.000}{2\pi\times2.2\times10^{-3}}[/tex]

[tex]r=5.23\times10^{-3}\ m[/tex]

Hence, The radius of the toroid is [tex]5.23\times10^{-3}\ m[/tex].