Bar Magnet in a Uniform Magnetic Field: Force, Torque, Potential Energy

Bar Magnet in a Uniform Magnetic Field: Do you know how the magnet could affect objects by attracting iron nails or pins placed nearby it? When you place a bar magnet in a uniform magnetic field, then the two poles of the bar magnet experience a force equal in magnitude and opposite in direction that does not have the same line of action. These forces constitute a couple that produces a turning effect called torque.

The torque tries to rotate the magnet to align it parallel to the direction of the field. In this article, we will learn about the bar magnet, the magnetic field lines, and the bar magnet as an equivalent solenoid. We will also discuss the torque and the potential energy of a bar magnet when placed in a uniform magnetic field.

Bar Magnet

When we sprinkle some iron filings on a sheet of glass placed over a short bar magnet, then these iron filings form some pattern which shows that the magnet has two poles similar to the positive and negative charge of an electric dipole; in which one pole is named as the North pole and the other as the South pole. When we suspend a bar magnetly, these poles point approximately towards the geographic north and south poles, respectively. Around a current-carrying solenoid, a similar pattern of iron filings is also observed.

Magnetic Field Lines

As in the figure below, the magnetic field lines of a bar magnet are shown. Some important properties of the magnetic field lines are given following:

1. The magnetic field lines form closed continuous loops, and also they do not intersect with each other.
2. The tangent to the field line at a given point gives the direction of the net magnetic field \(B\) at that point.
3. More the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field \(B\)

Bar Magnet as an Equivalent Solenoid

As per Ampere’s hypothesis, the magnetic dipole moment \((m)\) associated with a current loop is defined to be \({\rm{m}} = N/A\), where \(N\) is the number of turns in the loop, \(I\) the current, and the area vector\((A)\). Now, for the analogy that a bar magnet act as an equivalent solenoid, we calculate the axial field of a finite solenoid that resembles that of a bar magnet at large distances.

As shown in the figure below, let the solenoid consist of \(n\) turns per unit length with length \(\left( {2l} \right)\) and radius \(\left( a \right)\). And the axial field at a point, at a distance from the center \(\left( O \right)\) of the solenoid by considering a circular element of a thickness \(\left( {dx} \right)\) of the solenoid at a distance \(\;\left( x \right)\) from its center. It consists of \(n\;dx\) turns. Let assume that the current in the solenoid is \(I\).

Then, the magnitude of the magnetic field due to the circular element at point \(P\) is given as :

\(dB = \frac{{{\mu _0}ndxI{a^2}}}{{2{{\left[ {{{\left( {r – x} \right)}^2} + {a^2}} \right]}^{3/2}}}}\)

After approximation, the magnitude of the total field is obtained by summing over all the elements by integrating from \(x = \;–l\) to \(x =  + \;l\).

\(B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2m}}{{{r^3}}}\)

Hence, it shows that a bar magnet and a solenoid produce similar magnetic fields.

Torque on Magnetic Bar Placed in a Uniform Magnetic Field

The magnetic field lines give us an approximate idea of the magnetic field. To determine the magnitude of \(B\) accurately, let us consider a small compass needle of known magnetic moment \((m)\) and moment of inertia \(\left( {\cal I} \right).\) And then place this compass needle in the uniform magnetic field and allowing it to oscillate, as shown in the figure below.

As we know that :

\({\rm{\tau }} = {\rm{Magnetic\;Force \times perpendicular\;distance}}\)

Then, the torque on the needle can be expressed as:

\(\vec \tau = \vec m \times \vec B\)

Potential Energy of a Bar Magnet in a Uniform Magnetic Field

A bar magnet (magnetic dipole) of dipole moment \(\left( {\vec m\;} \right)\) is held at an angle \(\left( \theta  \right)\) with the direction of a uniform magnetic field \(\left( {\vec B} \right)\), as shown in the above figure. Then, we can write the magnitude of the torque acting on the dipole as:

\(\tau = mB\sin \theta \)

Here \(\tau \) is restoring torque and \(\theta \) is the angle between \(m\) and \(B\)

Therefore, in equilibrium:

\({\cal I}\frac{{{{\rm{d}}^2}\theta }}{{{\rm{d}}{t^2}}} =  – mB\sin \theta \)

A negative sign with \(mB\sin \theta \) implies that restoring torque is in opposition to deflecting torque. For small values of \(\theta \) in radians, we approximate \(\sin \theta  \approx \theta \) and get:

\({\cal I}\frac{{{{\rm{d}}^2}\theta }}{{{\rm{d}}{t^2}}} \approx  – mB\theta \)

or

\(\frac{{{{\rm{d}}^2}\theta }}{{{\rm{d}}{t^2}}} =  – \frac{{mB}}{{\cal I}}\theta \)

This represents a simple harmonic motion. The square of the angular frequency is:

\({\omega ^2} = \frac{{mB}}{{\cal I}}\)

And, the time period is,

\(T = 2\pi \sqrt {\frac{{\cal I}}{{mB}}} \)

or

\(B = \frac{{4{\pi ^2}{\cal I}}}{{m{T^2}}}\;\;\;\;\;\;\;\;.\;\;.\;\;.\;\;.\;\;\;\left( 1 \right)\)

Similar to electrostatic potential energy, an expression for magnetic potential energy can also be obtained on lines. The magnetic potential energy \({U_m}\) is given by:

\({U_m} = \smallint \tau \left( \theta  \right){\rm{d}}\theta \)

\( = \smallint mB\,{\rm{sin}}\theta \,{\rm{d}}\theta  =  – mB\cos \theta \)

\({U_m} =  – \vec m \cdot \vec B\;\;\;\;.\;\;.\;\;.\;\;.\;\;\;\left( 2 \right)\)

Now, taking the constant of integration to be zero means fixing the zero potential energy at \(\theta  = \;90^\circ ,\) i.e., when the needle is perpendicular to the field. Equation \((1)\) shows that potential energy is minimum \(\;\left( { =  – mB} \right)\) at \(\theta \; = \;0^\circ \) (most stable position) and maximum \(\left( { =  + mB} \right)\) at \(\theta \; = \;180^\circ \) (most unstable position).

Summary

The magnetic field lines form closed continuous loops, and also they do not intersect with each other. A bar magnet experiences no force when it is placed in a uniform magnetic field. The bar magnet experience a torque \(\vec \tau  = \vec m \times \vec B\). The direction of the torque is perpendicular to the plane containing \(\vec m\;\) and \(\vec B\). This torque produces the rotational motion of the magnet. When the dipole moment vector is perpendicular to the magnetic field, the magnet experiences maximum torque.

When the dipole moment vector is parallel or antiparallel to the magnetic field, in this case, the magnet experiences minimum torque. To get the potential energy of a bar magnet, we apply the dot product of the magnetic moment and the uniform magnetic field. When the magnet is perpendicular to the magnetic field, it has minimum potential energy and when it is aligned with it has maximum potential energy.

\({U_m} =  – \vec m \cdot \vec B\)

Learn the Concepts on Electromagnet

Bar Magnet in a Uniform Magnetic Field – Solved Examples

Q.1. A short bar magnet experiences a torque of \({\rm{0}}{\rm{.016\;}}\,{\rm{Nm}}\) when placed with its axis at \(30^\circ \) with an external field of \(800\;\,{\rm{G}}.\;\) Then, \((a)\) Find the magnetic moment of the magnet? \((b)\) From moving it from its most stable to a most unstable position, what is the work done? \((c)\) Find the current flowing through the solenoid when the bar magnet is replaced by a solenoid of cross-sectional area \(2 \times {10^{ – 4}}\;{{\rm{m}}^2}\) and \(1000\) turns, but of the same magnetic moment.
Sol: 
(a) From equation: \(\tau  = mB\;{\rm{sin}}\theta ,\;\theta  = 30^\circ \), hence \({\rm{sin}}\theta  = ½\)

Thus, \(0.016 = m \times \left( {800 \times {{10}^{ – 4\;}}{\rm{T}}} \right) \times \left( {1/2} \right)\)

\( \Rightarrow m = 160 \times 2/800 = 0.40\;\,{\rm{A}}{{\rm{m}}^2}\)

(b) From equation: \(\tau  = mB\;{\rm{sin}}\theta \)

\(\theta  = 0^\circ \) isthe most stable position and \(\theta  = 180^\circ \) is the most unstable position. The work done is

\(W = {U_m}\left( {\theta  = 180^\circ } \right) – {U_m}\left( {\theta  = 0^\circ } \right)\)

\( \Rightarrow W = 2mB = 2 \times 0.40 \times 800 \times {10^{ – 4}} = 0.064{\rm{\;J}}\)

(c) From the equation:  \({m_s} = NIA,\)  and given \({m_s} = 0.40\;{\rm{A\;}}{{\rm{m}}^2}\). Then

\(0.40 = 1000 \times I \times 2 \times {10^{ – 4}}\)

\( \Rightarrow I = \frac{{0.40 \times {{10}^4}}}{{1000 \times 2}}\)

\( \Rightarrow I = 2\;{\rm{A\;}}\)

FAQ’s on Bar Magnet in a Uniform Magnetic Field

Q. 1. What happens to the bar magnet when it is placed in a uniform magnetic field?
Ans: When a bar magnet is placed in a uniform magnetic field. Then the bar magnet will experience only torque and no force, and this torque on the bar magnet will be acting on both ends, will be equal but opposite in direction.

Q. 2. How are the magnetic field lines directed in a bar magnet?
Ans: A bar magnet has two magnetic poles: the south and the north poles, of equal pole strength near its ends. Outside the magnet, the magnetic field lines are directed from the north pole to the south pole, and inside the magnet, the magnetic field lines are directed from the south pole to the north pole.

Q. 3. What is the torque acting on a bar magnet?
Ans: A bar magnet experiences a torque when it is placed in a uniform magnetic field. A couple is a pair of two equal and opposite forces having different lines of action. They give rise to a turning effect known as torque. It is given as :
\(\vec \tau  = \vec m \times \vec B\)
Where \(\vec m\) is the magnetic moment and \(\vec B\) is the magnetic field.

Q. 4. What is the potential energy of a bar magnet in a uniform magnetic field?
Ans: The work done in rotating a bar magnet from a direction perpendicular to the field to any given direction is the potential energy of the bar magnet in a uniform magnetic field. We can get this by the dot product of the magnetic moment and the uniform magnetic field that is given as:
\({U_m} =  – \vec m \cdot \vec B\)
Where the magnetic moment is \(\vec m\), and the magnetic field is \(B\).

Q. 5. Give an example of magnetic potential energy?
Ans: Some items with high magnetic potential energy respond strongly to a magnetic force, such as metal spoons or ball bearings. Objects with lower magnetic potential energy are not as affected by a magnetic pull, such as a plastic spoon or a rubber ball.