Question

A toroidal solenoid (see the figure ) has inner radius 14.1cm and outer radius 18.6 cm . The solenoid has 270 turns and carries a current of 7.30 A. 

Part A 

What is the magnitude of the magnetic field at 11.8 cm from the center of the torus? 

Part B 

What is the magnitude of the magnetic field at 16.3 cm from the center of the torus? 

Part C 

What is the magnitude of the magnetic field at 20.4 cm from the center of the torus?

Answer 1

Concepts and reason

The concepts required to solve this question are toroidal magnetic field and Ampere’s law.

Firstly, find the magnetic field due to the toroid at a point inside the toroidal loop by using the ampere’s law for toroid.

Then, find the magnetic field inside the toroid by using the ampere’s law for toroid.

Finally, find the magnetic field outside the toroid circle by using the ampere’s law for toroid.

Fundamentals

The Ampere’s law states that the line integral of the magnetic field around some closed loop is equal to the product of the permeability of free space ( ${\mu _{\rm{o}}}$ ) and the current (I) enclosed by the loop.

$\oint {\vec B \cdot d\vec l} = {\mu _{\rm{o}}}I$

The magnetic field inside a toroid by using the Ampere’s law can be written as follows:

$B = \frac{{{\mu _{\rm{o}}}NI}}{{2\pi r}}$

Here, ${\mu _{\rm{o}}}$ is the permeability of the free space, N is the number of turns, I is the current in the toroid wire, and r is the distance between the center of the toroid circle and the point at which the magnetic field is to be found.

A)

The expression for the magnetic field can be written as follows:

$\oint {\vec B \cdot d\vec l} = {\mu _{\rm{o}}}I$

For a closed loop of radius less than the inner radius:

Consider a loop of radius r which is inside the toroid and has a radius less than the inner radius of the toroid.

There is no source of current within this loop so that the current enclosed by the loop is zero.

By Ampere’s circuital law,

$\oint {\vec B \cdot d\vec l} = {\mu _{\rm{o}}}I$

Substitute 0 for I in the above expression.

$\oint {\vec B \cdot d\vec l} = 0$

The line element cannot be zero so that the only quantity which can be zero to make the above expression true is magnetic field.

Thus, $\vec B = 0$ .

B)

The magnetic field inside a toroid by using the Ampere’s law can be written as follows:

$B = \frac{{{\mu _{\rm{o}}}NI}}{{2\pi r}}$

Substitute $4\pi \times {10^{ - 7}}{\rm{ H/m}}$ for ${\mu _{\rm{o}}}$ , 270 for N, 16.3 cm for r, and 7.30 A for I in the above expression.

$\begin{array}{c}\\B = \frac{{\left( {4\pi \times {{10}^{ - 7}}{\rm{ H/m}}} \right)\left( {270} \right)\left( {7.30{\rm{ A}}} \right)}}{{2\pi \left( {16.3{\rm{ cm}}} \right)}}\left( {\frac{{1{\rm{ cm}}}}{{{{10}^{ - 2}}{\rm{ m}}}}} \right)\\\\ = 0.0024{\rm{ T}}\\\end{array}$

C)

The expression for the magnetic field can be written as follows:

$\oint {\vec B \cdot d\vec l} = {\mu _{\rm{o}}}I$

For a closed loop of radius greater than the outer radius:

Consider a loop of radius r which is outside the toroid and has a radius greater than the outer radius of the toroid.

The current within this loop cancels out so that the current enclosed by the loop is zero.

By Ampere’s circuital law,

$\oint {\vec B \cdot d\vec l} = {\mu _{\rm{o}}}I$

Substitute 0 for I in the above expression.

$\oint {\vec B \cdot d\vec l} = 0$

The line element cannot be zero so that the only quantity which can be zero to make the above expression true is magnetic field.

Thus, $\vec B = 0$ .

Ans: Part A

The magnetic field is zero.

Part B

The magnetic field is 0.0024 T.

Part C

The magnetic field is zero.