Question

a toroid of narrow radius of 10 cm has 1000 turns of wire. for a magnetic field of 5*10^-2 along its axis,how much current is required to be passed the through the wire ?
​

Answer 1

Given:

A toroid of narrow radius of 10 cm has 1000 turns of wire. for a magnetic field of 5*10^-2 along its axis.

To find:

Current to be passed through the coils.

Calculation:

General expression for magnetic field in a toroid is given as follows :

$\boxed{ \bold{B = \dfrac{ \mu_{0} Ni}{2\pi r}}}$

Putting available values in SI units:

$\rm{ = > B = \dfrac{2 \mu_{0} Ni}{4\pi r}}$

$\rm{ = > B = \dfrac{ \mu_{0} }{4\pi} \times \dfrac{ 2Ni}{r}}$

$\rm{ = > B = {10}^{ - 7} \times \dfrac{ 2 \times 1000 \times i}{ \frac{10}{100} }}$

$\rm{ = > 5 \times {10}^{ - 2} = {10}^{ - 7} \times \dfrac{ 2 \times 1000 \times i}{ {10}^{ - 1} }}$

$\rm{ = > 5 \times {10}^{ 5} = \dfrac{ 2 \times 1000 \times i}{ {10}^{ - 1} }}$

$\rm{ = > 5 \times {10}^{ 2} = \dfrac{ 2 \times i}{ {10}^{ - 1} }}$

$\rm{ = > 5 \times {10}^{ 2} = 20 \times i}$

$\rm{ = > i = \dfrac{500}{20} }$

$\rm{ = > i = 25 \: amp}$

So, final answer is:

$\boxed{ \large{ \bf{ i = 25 \: amperes}}}$