Question

A current of 16.0 mA is maintained in a single circular loop of 1.90 m circumference. A magnetic field of 0.790 T is directed parallel to the plane of the loop.

(a) Calculate the magnetic moment of the loop.
mA · m²

(b) What is the magnitude of the torque exerted by the magnetic field on the loop?
mN · m

Answer 1

Concepts and reason

The main concepts required to solve this problem are the magnetic moment, current, magnetic field, area and torque.

Initially, write the equation for the magnetic moment of the loop, area of the loop and the toque exerted by the magnetic field in the loop. Use these equations and calculate the magnetic moment of the loop and the toque exerted by the magnetic field in the loop.

Fundamentals

The equation for the magnetic moment of the loop is,

$\mu = IA$

Here, I is the current through the loop, and A is the area of the loop.

The equation for the area of the loop is,

$A = \pi {r^2}$

Here, r is the radius of the loop.

The equation for the circumference of the circular loop is,

${\rm{C}} = 2\pi r$

Here, r is the radius of the loop.

The equation for the torque exerted by the magnetic field in the loop is,

$\tau = IAB$

Here, I is the current, A is the area of the loop and B is the magnetic field.

(a)

The equation for the circumference of the circular loop is,

${\rm{C}} = 2\pi r$

Rearrange the above equation for r.

$r = \frac{{\rm{C}}}{{2\pi }}$

Substitute 1.90 m for C, and 3.14 for $\pi$ in above equation.

$\begin{array}{c}\\r = \frac{{1.90{\rm{ m}}}}{{2\left( {3.14} \right)}}\\\\ = 0.3025{\rm{ m}}\\\end{array}$

The equation for the area of the loop is,

$A = \pi {r^2}$

Substitute 3.14 for $\pi$ , and $0.3025{\rm{ m}}$ for r in above equation.

$\begin{array}{c}\\A = \left( {3.14} \right){\left( {0.3025{\rm{ m}}} \right)^2}\\\\ = 0.287{\rm{ }}{{\rm{m}}^2}\\\end{array}$

The equation for the magnetic moment of the loop is,

$\mu = IA$

Substitute $16.0{\rm{ mA}}$ for I and $0.287{\rm{ }}{{\rm{m}}^2}$ for A in above equation.

$\begin{array}{c}\\\mu = \left( {16.0{\rm{ mA}}} \right)\left( {0.287{\rm{ }}{{\rm{m}}^2}} \right)\\\\ = 4.58\,\,{\rm{mA}} \cdot {{\rm{m}}^2}\\\end{array}$

(b)

The equation for the torque exerted by the magnetic field in the loop is,

$\tau = IAB$

Substitute $16.0{\rm{ mA}}$ for I, $0.287{\rm{ }}{{\rm{m}}^2}$ for A, and $0.790{\rm{ T}}$ for B in above equation.

$\begin{array}{c}\\\tau = \left( {16.0{\rm{ mA}}} \right)\left( {0.287{\rm{ }}{{\rm{m}}^2}} \right)\left( {0.790{\rm{ T}}} \right)\\\\ = 3.62{\rm{ mN}} \cdot {\rm{m}}\\\end{array}$

Ans: Part a

The magnetic moment of the loop is $4.58\,\,{\rm{mA}} \cdot {{\rm{m}}^2}$ .

Part b

The torque exerted by the magnetic field in the loop is $3.62{\rm{ mN}} \cdot {\rm{m}}$ .