Dipole Uniform Magnetic Field

A magnetic dipole consists of two equal and opposite magnetic poles separated by a distance. Bar magnets, current-carrying loops, and certain atomic and subatomic particles exhibit magnetic dipole behavior. The magnetic dipole moment points from the south to the north pole of a bar magnet or the direction given by the right-hand rule for current loops. Magnetic dipoles are fundamental in the operation of compasses and various magnetic sensors.

1.0Uniform Magnetic Field

A magnetic field is considered uniform if it maintains both constant strength and direction throughout the entire region.

• The magnitude and direction of the magnetic field strength remain constant throughout the region.
• Uniform magnetic field is represented by an equidistant set of parallel lines.

Examples of Uniform Magnetic Field:

• Certain Magnetic Field Generators
• Magnetic Field Between Poles of a Solenoid

2.0Non Uniform Magnetic Field

A magnetic field is non-uniform if its strength varies at different points within the region.

• The magnitude and direction of the magnetic field strength change at different points within the region..
• Non Uniform Magnetic Field is represented by converging, diverging, or un-equispaced field lines.

NOTE:A strong magnetic field is represented by closely spaced lines whereas weak magnetic field is represented by widely spaced lines.

Examples of Non Uniform Magnetic Field:

• Magnetic Field of a Current Loop
• Magnetic Field of a Bar Magnet

3.0Dipole in a Uniform Magnetic Field

A magnetic dipole consists of a pair of magnetic poles of equal and opposite strength separated by a small distance. 

Examples: Magnetic needle, Bar Magnet

4.0Magnetic Dipole Moment

• It is defined as the product of the pole strength of either pole and the distance between the magnetic poles
• $\vec{M}=\mathrm{m}\vec{l}$
• Number of field lines emerging from each pole indicate the pole strength of the pole.
• Pole strength m of each pole ∝ area of cross section for a given bar magnet.
• Pole Strength Unit- Ampere–metre
• $\vec{M}$ is a vector quantity
• Direction of the magnetic dipole moment is from the south to north pole.
• $\text{ SI Unit-Ampere-metre }{}^2\left(A{m}^2\right)$
• $\text{ Dimensional Formula- }\left[M{L}^2{T}^{0}A\right]$
• A magnetic moment is a vector, in the case of two magnets having magnetic moments M₁ and M₂ with angle between them, the resulting magnetic moment,

$M={\left[M_1^2+M_2^2+2M_1{M}_2\mathrm{Cos}\theta \right]}^{\frac{1}{2}}$

$\mathrm{Tan}\varphi =\left[\frac{M_2\mathrm{Sin}\theta }{M_1+M_2\mathrm{Cos}\theta }\right]$

5.0Torque on Magnetic Dipole in Uniform Magnetic Field

When a bar magnet or a coil with a dipole moment M is placed in a uniform magnetic field B, it experiences a torque that acts to align the dipole moment with the direction of the external magnetic field. Work has to be done in rotation of the magnet or coil.

1. Coil or Loop

If a loop carries a current of magnitude I and magnetic moment $\vec{M}$ held in uniform magnetic field $\vec{B}$ making an angle ,then the loop experiences a torque given by

$\vec{\tau }=\vec{M}\times \vec{B}$

$\tau =MB\sin \theta$

$\vec{\tau }=I(\vec{A}\times \vec{B})$

If loop have N turns then 

$\vec{\tau }=NI(\vec{A}\times \vec{B})$

$\tau =N\mathrm{IAB}\mathrm{Sin}\theta$ a uniform magnetic field $\vec{B}$ making an angle \theta with the field.Then the magnet experiences a torque is given by,

$\tau =\text{ Force }\times \text{ Perpendicular distance between force couple }$

$\tau =F(l\mathrm{Sin}\theta )$

$\tau =(mB)(l\mathrm{Sin}\theta )$

$\tau =(ml)B\mathrm{Sin}\theta$

$\tau =MB\mathrm{Sin}\theta \text{ where }M=ml$

$\vec{\tau }=\vec{M}\times \vec{B}$

$\tau =MB\mathrm{Sin}\theta$

$\Rightarrow 90^{\circ }\Rightarrow \tau =MB(\text{ maximum })$

$\Rightarrow 0^{\circ }\text{ or }180^{\circ }\Rightarrow \tau =0\text{ (minimum) }$

NOTE:

• Torque on a dipole is an axial vector and it is directed along the axis of rotation of the dipole.
• Tendency of torque on dipole is to try to align the  in the direction of  or tries to make the axis of dipole parallel to  or make the plane of coil (or loop) perpendicular to .
• Dipole in uniform magnetic field

$\Rightarrow F_{\text{net }}=0(\text{ no Translatory motion })$

$\Rightarrow \tau \text{ may or may not be zero decides by }\theta$

• Dipole in Non-Uniform magnetic field

$\Rightarrow F_{\text{net }}\text{ may or may not be zero }$

$\Rightarrow \tau \text{ may or may not be zero decides by }\theta$

6.0Potential Energy Of Magnetic Dipole 

1. Work Done in Rotating the Coil in Uniform Magnetic Field

When a bar magnet or coil of the magnetic dipole moment M is placed in a uniform magnetic field B it experiences a torque which begins to align it in the direction of external magnetic field $\vec{B}$ Work has to be done in rotation of the magnet or coil.

The work done in rotating dipole by small angle $d\theta \text{ is }dW=\tau d\theta$

The work done in rotating the magnet or coil from $\theta ={\theta }_1\text{ to }\theta ={\theta }_2$ is given by

$W={\int }_{{\theta }_1}^{{\theta }_2}\tau d\theta$

$W={\int }_{{\theta }_1}^{{\theta }_2}MB\sin \theta d\theta$

$W=MB[-\mathrm{Cos}\theta {]}_{{\theta }_1}^{{\theta }_2}$

$W=-MB\left(\mathrm{Cos}{\theta }_2-\mathrm{Cos}{\theta }_1\right)$

$W=MB\left(\mathrm{Cos}{\theta }_1-\mathrm{Cos}{\theta }_2\right)$

2. Potential Energy of the coil in Uniform Magnetic Field

It is defined as work done in rotating the dipole from a direction perpendicular to the given direction.

Work done in rotating the dipole from ${\theta }_1=90^{\circ }\text{ to }{\theta }_2=0^{\circ }$

$W=MB\left(\mathrm{Cos}90^{\circ }-\mathrm{Cos}\theta \right)$

$U=-MB\mathrm{Cos}\theta$

$U=-\vec{M}\cdot \vec{B}$

Important Note:

Case-1	Case-2	Case-3
When $\vec{M}$ and $\vec{B}$ are parallel, $\theta =0^{\circ }$ $\tau =0$ $U_{\min }=-MB$ The dipole has minimum potential energy and it is in stable equilibrium.	When $\vec{M}$ and $\vec{B}$ are Perpendicular $\theta =90^{\circ }$ $\tau =MB$ U=0 No Equilibrium as $\tau \text{ is not equal to zero }$	When $\vec{M}$ and $\vec{B}$ are anti parallel $\begin{array}{rl}& \theta =180^{\circ }\\ & \tau =0\\ & U_{\max }=MB\end{array}$ The dipole has maximum potential energy and it is in Unstable equilibrium.

7.0Sample Questions on Dipole Uniform Magnetic Field

Q-1.Two wires of equal length are shaped into two loops: one is square and the other is circular. Both loops are suspended in a uniform magnetic field and carry the same current. Which loop will experience a greater torque?

Solution: Torque acting on a loop, $\tau =NIAB$ Since the area of the circular loop is greater than that of the square loop, the circular loop will experience a larger torque.

Q-2.Explain angular SHM of Magnetic Dipole?

Solution: When a dipole is suspended in a uniform magnetic field it will align itself parallel to the field. Now if it is given a small angular displacement $\theta$ with respect to its equilibrium position. The restoring torque acts on it ,

$\tau =-MB\mathrm{Sin}\theta$

$\text{ For small angle, }\mathrm{Sin}\theta \approx \theta$

$\tau =-MB\theta$

For angular SHM, $\tau =-I{\omega }^2\theta$

$-I{\omega }^2\theta =-MB\theta$

$\omega =\sqrt{\frac{MB}{I}=\frac{2\pi }{T}}$

$T=2\pi \sqrt{\frac{I}{MB}}\text{ where }I=\text{ Moment of Inertia }=\frac{m{l}^2}{12},(m\text{ mass of magnet, }l:\text{ length of magnet })$

Q-3.A bar magnet with a length of 10 cm and a pole strength $10^{-3}$ A-m is placed in a magnetic field B $4\pi \times 10^{-3}T$. The magnet is oriented at an angle of 30° to the direction of the magnetic field. Calculate the torque acting on the magnet.

Solution:

$\tau =MB\mathrm{Sin}\theta =0.1\times 10^{-3}\times 4\pi \times 10^{-3}\times \mathrm{Sin}30^{\circ }=10^{-7}\times 4\pi \times \frac{1}{2}=2\pi \times 10^{-7}\mathrm{Nm}$

Q-4.Find the net magnetic moment of two identical magnets each of magnetic moment $m_0$ inclined at $60^{\circ }$ with each other?

Solution:

$M={\left[M_1^2+M_2^2+2M_1{M}_2\mathrm{Cos}\theta \right]}^{\frac{1}{2}}$

$M=\sqrt{m_0^2+m_0^2+2m_0{m}_0\mathrm{Cos}60^{\circ }}$

$M=\sqrt{3m_0^2}=\sqrt{3}{m}_0$

Q-5. A magnetic needle aligned parallel to a magnetic field requires W units of work to rotate it through 60°.  Determine the torque needed to keep the needle in this position.

Solution:

$\begin{array}{c}W=MB\left(\mathrm{Cos}{\theta }_1-\mathrm{Cos}{\theta }_2\right)\\ {\theta }_1=0^{\circ }\text{ and }{\theta }_2=60^{\circ }\\ W=MB\left(\mathrm{Cos}{0}^{\circ }-\mathrm{Cos}60^{\circ }\right)\\ W=MB\left(1-\frac{1}{2}\right)=\frac{MB}{2}\\ MB=2W\end{array}$

Restoring torque acting on the needle is $\tau =MB\mathrm{Sin}\theta$ 

$\tau =2W\mathrm{Sin}60^{\circ }=2W\times \frac{\sqrt{3}}{2}=\sqrt{3}\mathrm{W}$