Question

A closely wound solenoid of 750 turns and area of cross-section of $5\times {{10}^{-4}}{{m}^{2}}$ carries a current of $3.0A$. Its associated magnetic moment is?
A. $4.12\text{ J}{{\text{T}}^{-1}}$
B. $3.12\text{ J}{{\text{T}}^{-1}}$
C. $2.12\text{ J}{{\text{T}}^{-1}}$
D. $1.13\text{ J}{{\text{T}}^{-1}}$

Answer 1

Hint: The solenoid consists of a helically shaped coil of many turns and a long length. The magnetic moment of a coil is defined as the product of the current flowing through the coil and the area of the coil. Since the solenoid consists of a number of terms, all these terms will contribute to the magnetic moment.

Complete Step-by-Step solution:
In the given problem, the number of turns is 750, area of cross-section of the solenoid is $5\times {{10}^{-4}}{{m}^{2}}$ and the current flowing through the solenoid is $3.0A$.
So the magnetic moment associated with an object carrying a current i is given by, \[\text{Magnetic Moment}(\mu )=i\overrightarrow{A}\]. So if there are N turns associated with a coil like in a solenoid, each coil will produce a magnetic moment and the total magnetic moment can be written as,
\[\text{Magnetic Moment}(\mu )=Ni\overrightarrow{A}\]
So substituting the values of current, area of cross section and number of turns into the above equation, we get,
\[\text{Magnetic Moment}(\mu )=\left( 750 \right)\times \left( 3.0A \right)\times \left( 5\times {{10}^{-4}}{{m}^{2}} \right)\]
$\therefore \mu =1.125J{{T}^{-1}}\approx 1.13J{{T}^{-1}}$
So the magnetic moment of a solenoid which is directed along the axis has a magnitude $\mu \approx 1.13J{{T}^{-1}}$.
So the answer to the question is option (D).

Note: The SI unit for the magnetic moment is $Ampere-{{(meter)}^{2}}$. There can be other derived units by the making use of other units. We can express magnetic moment by the torque applied per unit magnetic field. So we can write,
$\text{Magnetic Moment}=\dfrac{\text{Torque}}{\text{Magnetic Field}}$
Writing out the units we can see that,
$\text{Magnetic Moment}=\dfrac{Force.meter}{Tesla}$
$\text{Magnetic Moment}=\dfrac{N.m}{T}$
So the unit of magnetic moment is given by $\dfrac{N.m}{T}$
We know that the product of force and displacement is the work done on the magnetic field by the magnetic field, which is stored as magnetic potential energy inside the field. So we can write the unit of the magnetic moment as $\dfrac{Joule}{Tesla}\text{ or }\dfrac{J}{T}$. This is the unit used in the problem.