A circular current loop of magnetic moment M is in an arbitrary orientation in an external magnetic field \$\bar B\$. The work done to rotate the loop by 30 about an axis perpendicular to its plane is:

The rotation of the loop by \$30°\$ about an axis perpendicular to its plane make no change in the angle made by axis of the loop with the direction of magnetic field, therefore, the work done to rotate the loop is zero.Important point: The work done to rotate the loop the loop in magnetic field \$W = MB(cos\ \theta_1 - cos\ \theta_2)\$, where signs are as usual.

A particle of charge q and mass m moves in a circular orbit of radius r with angular speed \$\omega\$. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on

The angular momentum L of the particle is given by \$\displaystyle L=mr^{2}w\$ where \$w=2\pi n\$.\$\therefore\$ Frequency \$\displaystyle n=\frac{w}{2\pi }\$;Further \$\displaystyle i= q\times n=\frac{wq}{2\pi}\$Magnetic Moment, \$\displaystyle M=iA=\frac{wq}{2\pi}\times \pi r^{2}\$;\$\displaystyle \therefore M=\frac{wqr^{2}}{2}\$So, \$\displaystyle \frac{M}{L}=\frac{wqr^{2}}{2mr^{2}w}=\frac{q}{2m}\$Thus the required ratio depends upon both q and m.

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Dipole in Uniform Magnetic Field

Q: Derive an expression for the magnetic dipole moment of a revolving electron.

The current in a loop due to a revolving electron \$=\dfrac{q}{T}\$ \$=\dfrac{q}{2\pi r/v}\$ \$=\dfrac{qv}{2\pi r}\$Hence, magnetic moment due to the revolving charge \$=IA\$ \$=\dfrac{qv}{2\pi r}\pi r^2\$ \$=\dfrac{qvr}{2}\$

Q: Obtain an expression for the orbital magnetic moment of an electron rotating about the nucleus in an atom.

Consider an electron revolving around a nucleus in an atom. Since the electron is a negatively charged particle, so atom will act as a current-carrying loop and its magnetic dipole moment is given by \$ M=lA \$Where \$ l\$ =current and \$ A \$ =area of cross-sectionif \$ r \$ is the radius of orbit then\$ M=\left( \frac { e }{ t } \right) \left( \pi r^{ 2 } \right) \$=\$ \frac { e }{ \frac { 2\pi r }{ v } } \pi r^2=\frac{evr}{2} \$Sice \$ mvr\$ = angular momentum \$ (L) \$So, \$ M=\left( \frac { 2 }{ 2m_{ e } } \right) L \$ (in vector form), since magnetic dipole moment of the electron and angular momentum are in opposite direction.

Q: A charged particle (charge q) is moving in a circle of radius R with uniform speed v. The associated magnetic moment \$\mu\$ is given by

The current in circular path due to movement in circle is given by \$i=\dfrac{q}{T}=\dfrac{qv}{2\pi R}\$Hence the magnetic moment associated with the particle is \$\mu=iA=\dfrac{qv}{2\pi R}\pi R^2\$\$=\dfrac{qvR}{2}\$

Q: A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. the energy required to rotate it by \$60^o\$ is \$W\$, Now the torque required to keep the magnet in this new position is.

Torque acting on a bar magnet kept in a magnetic field \$\vec{B}\$ is \$\tau=\vec{m}\times \vec{B}=mBsin\theta\$Hence work done in moving from \$\theta=0^{\circ}\$ to \$\theta=60^{\circ}\$ is \$W=mBcos 60^{\circ}=\dfrac{1}{2}mB\$Hence torque acting when magnet is placed at \$60^{\circ}=\dfrac{\sqrt{3}}{2}mB=\sqrt{3}W\$

Q: Write the expression for magnetic potential energy of a magnetic dipole kept in a uniform magnetic field and explain the terms.

\$U=-\vec { m } .\vec { B } \quad or\quad U=mB\cos { \theta } \$\$U-\$ Magnetic potential energy\$m-\$ magnetic moment of the magnetic dipole\$B-\$ Uniform magnetic field

Q: A circular loop of area \$1\$ \$cm^2\$, carrying a current of \$10\$A, is placed in a magnetic field of \$0.1\$T perpendicular to the plane of the loop. The torque on the loop due to the magnetic field is?

As Torque is given by\$T = NBiA\sin \theta\$Area \$= 1cm^2\$Current \$= 10A\$Magnetic field \$= 0.1T\$\$\theta = 0\$\$T= 1\times 10\times 0.1\times \sin 0\$ \$= 1 \times 0\$\$T = 0\$

Q: A short bar magnet of magnetic moment \$0.4 J T^{1}\$ is placed in a uniform magnetic field of 0.16 T. The magnet is in stable equilibrium when the potential energy is

Potential energy of a magnet with magnetic moment \$\vec{M}\$ placed in a magnetic field \$\vec{B}\$ is given by \$U=-\vec{M}.vec{B}\$\$=-MBcos\theta\$which is smallest for \$\theta=0\$Hence \$U_{min}=-MB=-0.4\times 0.16J\$\$=-0.064J\$

Q: A bar magnet of magnetic moment \$1.5 J T^{-1}\$ lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)?

Here \$m=1.5J/T, B=0.22T\$(a) \$W=-mB(cos\theta_2-cos\theta_1)\$ (i) \$\theta_1=0^{\circ}\$(along the field) \$\theta_2=90^{\circ}\$(perpendicular to the field) \$W=-0.33(0-1)J=0.33\ J\$ (ii) \$\theta_1=0^{\circ},\theta_2=180^{\circ}\$ \$W=-1.5\times 0.22(cos180^{\circ}-cos0^{\circ})=0.66\ J\$(b)Torque \$=mBsin\theta\$ (i) \$\theta=90^{\circ}\$ \$\implies \tau=0.33\ Nm\$ (ii) \$\tau=mBsin180^{\circ}\$ \$=0Nm\$

Physics

example

Potential energy of a magnet in a magnetic field

If a bar magnet in magnetic moment

m

is deflected from an angle

θ

in a uniform magnetic field of induction

B

, the energy stored in reversing the direction is :
Now lets assume that the bar magnetic is initially placed parallel to the field the initial potential energy will be

U_{i} = - M . B = - M B cos 0 = - M B

Now when it is deflected by an angle

θ

then

U_{f} = - M . B = - M B cos θ

Δ U = U_{i} - U_{f}

Δ U = - M B (1 - cos θ)

example

Potential energy of a current loop in a magnetic field

A planar coil of area

7

m^{2}

carrying an anti-clockwise current

2 A

is placed in an extemal magnetic field

B = (0.2 \hat{i} + 0.2 \hat{j} - 0.3 \hat{k})

, such that the normal to the plane is along the line

(3 \hat{i} - 5 \hat{j} + 4 \hat{k})

Area

A = 7 m^{2},

Normal

\overset{n}{^} = \frac{1}{5 0} (3 \hat{i} - 5 \hat{j} + 4 \hat{k})

≃ \frac{1}{7} (3 \hat{i} - 5 \hat{j} + 4 \hat{k})

∴

area

A = (3 \hat{i} - 5 \hat{j} + 4 \hat{k}) m^{2}

Magnetic field

B = (0.2 \hat{i} + 0.2 \hat{j} - 0.3 \hat{k}) T

Current

I = 2 A

∴

magnetic moment

M = I A = (6 \hat{i} - 10 \hat{j} + 8 \hat{k}) A m^{2}

∣ M ∣ ≃ 14 A m^{2} (2 \times 7 = 14)

Potential energy U

= - M . B

= - [1.2 - 2 - 2.4] = 3.2 J

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