Intensity of magnetic field and torque on a bar magnet - class-XII

Magnetic induction due to a short bar magnet on its axial line,

$B \displaystyle = \frac{\mu _0  2M}{4 \pi  d^3}$
At $\displaystyle \frac{d}{4} $ distance,
$B' \displaystyle = \frac{\mu _0 2M}{4 \pi (d/4)^3}$
$\displaystyle = \frac{\mu _0  2M}{4 \pi d^3} \times 64 = 64 B$

For a short Bar Magnet, the magnetic induction at a point on the axix at a distance $r$ from centre is given by  the formula

$r = \sqrt{13} d ;  2l  = 2d$
$B _eq = \displaystyle \frac{\mu _0}{4 \pi} \times \frac{\mu}{(r^2 + 1^2)^{\frac{3}{2}}}$
$\displaystyle =\frac{\mu _0}{4\pi} \times \frac{M}{\left ((\sqrt{13}d)^2 + (2d)^2 \right )^{\frac{3}{2}}}$
$=\displaystyle \frac{\mu _0}{4\pi} \times \frac{M}{(17 d^2)^{\frac{3}{2}}} = \frac{\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$

$B _{axial} = \displaystyle \frac{\mu _0}{4\pi} \frac{2Mr}{(r^2-l^2)^2}$
$B _{20} : B _{30} = 128  :  27$
$\displaystyle \frac{20}{(20^2 - l^2)^2} \times \frac{(30^2 - l^2)^2}{30} = \frac{128}{127}$
$2 (30^2 - l^2)^2 (27) = 3 (20^2 - l^2)^2 128$
$\sqrt{54}(900 - l^2) = \sqrt{384} (400 - l^2)$
$900 \sqrt{54} - \sqrt{54}l^2 = \sqrt{384} \times 400 - \sqrt{384}l^2$
$l^2 (12.2474) = 12224. 75$
$l^2 = 100 ;  l = \sqrt{100}-10 cm$
$\therefore$ Magnetic length $=2l = 20 cm$

$B _{equi} = \displaystyle \frac{\mu _0}{4 \pi } \frac{\mu}{r^3} = \frac{10^{-7}\times 5.4}{(0.3)^3} = 2 \times 10^{-5}T$

$B = \displaystyle \frac{\mu _0  2M}{4 \pi  d^3}$
At $(4d/5) $ distance
$B' = \displaystyle \frac{\mu _0  2M}{4 \pi \left ( \frac{4d}{5} \right )^3} = \frac{\mu _0  2M}{4  \pi (d^3)} \left ( \frac{125}{64} \right )$
$=\displaystyle \frac{125}{64}B$

When the north pole of short bar magnet is facing North pole of the earth, at the neutral point P, which is on equatorial line. 
$B _H = \dfrac {\mu _0 M}{4\pi d^3} = B _1$ ............(1)
When the magnet is rotated by $90^o$, the magnetic induction at P which is on axial line,
$B _H = \dfrac {\mu _0 2M}{4\pi d^3} = B _2$ ............(2)
Therefore, net magnetic induction at P is
$B _{net} = \sqrt {(B _1^2 + B _2^2)}$
$B _{net} = \sqrt {(1^2 + 2^2)}B _H = \sqrt 5 B _H$

The magnetic moment of the solenoid is M = N * I * A = 2000 * 2.0 * 1.5 * 10^-4 = 0.6 Am^2. The torque in a uniform magnetic field B is given by tau = M * B * sin(theta), where theta is the angle between the magnetic field and the axis of the solenoid. Thus, tau = 0.6 * (5 * 10^-2) * sin(30 deg) = 0.6 * 0.05 * 0.5 = 1.5 * 10^-2 Nm.

$T _1= 6 = 2\pi \sqrt{\dfrac{1}{(M _1+M _2)B}}$
$T _2= 12 = 2\pi \sqrt{\dfrac{1}{(M _1+M _2)B}}$
$\therefore \dfrac{6}{12}=\dfrac{1}{2}= \sqrt{\dfrac{M- M _2}{M _1+M _2}}$ or $ M _1+M _2= 4( M _1-M _2) $
$ 3M _1= 5M _2$
$\dfrac{M _1}{M _2}=\dfrac{5}{3}$

Maximum voltage (EMF) is given by E_max = N * B * A * omega. Frequency f = 0.5 rev/s, so omega = 2 * pi * f = 2 * pi * 0.5 = pi rad/s. E_max = 100 * 0.01 * 0.10 * pi = 0.1 * pi = 0.314 V.

The magnetic dipole moment of the bar magnet is M = pole strength * length = (500 * 10^-3 Am) * (0.16 m) = 0.08 Am^2. The external magnetic field B = 2.5 gauss = 2.5 * 10^-4 Tesla. Using the torque formula tau = M * B * sin(theta), we have sqrt(3) * 10^-5 = 0.08 * (2.5 * 10^-4) * sin(theta) = 2 * 10^-5 * sin(theta). Solving for sin(theta) gives sin(theta) = sqrt(3) / 2, which means theta = pi / 3.

Torque acting on the bar magnet suspended in a uniform field B is given by,

Torque is vector product of magnetic moment and the magnetic field.

$\tau = \bar M \times \bar B.$ since perpendicular, $\theta = 90^{\circ}$

A magnetic needle suspended freely orientates itself in a definite direction.

Given,

Magnetic field $\vec{B}\,=\,0.5\hat{i}\,+\,3.0\hat{j}\,\,T$

Magnetic moment $\mu =50\hat{i}\,\,A{{m}^{2}}$

Torque on magnet  $\tau $

$\tau =\mu \times \vec{B}\,\,=\,\,50\hat{i}\,\times \,(0.5\hat{i}\,+\,3.0\hat{j}\,)$

$\tau =\,150\hat{k}\,\,N-m$

Hence, Torque experiences by the magnet present in the magnetic field, $\tau =\,150\hat{k}\,\,N-m$ 

$\tau=MXB=MBsinQ$

Given that,

Dipole moment $m=10\left( \hat{i}+\hat{j}+\hat{k} \right)$

Magnetic field $B=0.6\hat{i}+0.4\hat{j}+0.5\hat{k}$

We know that,

The force is

  $ \tau =m\times B $

 $ \tau =\left( 10\hat{i}+10\hat{j}+10\hat{k} \right)\times \left( 0.6\hat{i}+0.4\hat{j}+0.5\hat{k} \right) $

 $ \tau =\hat{i}-\hat{j}-2\hat{k} $

Hence, the force acting on the dipole is $\hat{i}-\hat{j}-2\hat{k}$

From the given question, 

Given,

For a short magnet, magnetic intensity H at broad side-on (equatorial) is H1 = M / (4 * pi * d^3) and at end side-on (axial) is H2 = 2 * M / (4 * pi * d^3). The ratio H1:H2 = 1:2.

Magnetic moment M = pole strength (m) * length (L). Force on each pole F = m * B. Given M = 80, B = 1.8e-5, F = 25e-3. m = F / B = 25e-3 / 1.8e-5 = 25000 / 18 = 1388.89. L = M / m = 80 / 1388.89 = 0.0576 m = 5.76 cm.

Torque at an angle $\theta  = k sin \theta $

Average of the two field strengths at the poles is $\frac{0.2 +0.22}{2}=0.21$
Copuling( torque): $MB=4 \times 0.21 : Nm= 0.84:Nm$

$l=30\ cm$
$P=10\ Am$
$B=40\times 10^{-6}$
$\theta =60^o$
$m=pl$
$\vec{\tau}=\vec{m}\times \vec{B}$
$=mB\sin\theta $
$=10^{-6}\times \dfrac{\sqrt{3}}{2}$
$=1.039\times 10^{-4}Nm$
$=10.39\times 10^{-5}Nm$

The force between magnetic poles is F = k(m1m2)/d^2. If m1 becomes 2m1, to keep F constant, the new distance d' must satisfy (2m1m2)/(d')^2 = (m1*m2)/d^2. This simplifies to 2/d'^2 = 1/d^2, so d' = d/sqrt(2).

A magnetic needle is a magnetic dipole.
In a non uniform Magnetic field Force on each of the poles will be different in both magnitude and direction.
Due to difference in Magnitude the dipole experiences a Force,
Due to difference in Direction the dipole experiences a Torque.

$W = MB(cos\theta _1 cos \theta _2) = MB (cos 0^o cos 60^o)$

$=MB\left (1-\dfrac {1}{2}\right )=\dfrac {MB}{2}$

$\Rightarrow MB=2\sqrt 3J$

$\tau=MB sin 60^o=(2\sqrt 3)\left (\dfrac {\sqrt 3}{2}\right )J=3J$

Torque, $\tau = 0.64\ J, B = 0.32\ T, \theta = 30^{\circ}$
Torque, $\tau = MB\sin \theta$
$0.64 = M\times 0.32\sin 30^{\circ}$
$0.64 = M\times 0.32\times \dfrac {1}{2}$
$M = \dfrac {2\times 0.64}{0.32} = 4\ Am^{2}$.

Torque experienced by a magnet suspended in a uniform magnetic field B is given by
$\tau =MB \sin \theta$
Here, $M=200A m^2, B=0.30N A^{-1}m^{-1}$ and $\theta =30^o$
$\therefore \tau =200\times 0.30\times \sin 30^o$
$\tau =30$N m

The torque acting on bar magnet is given as,

$\tau  = M \times B$

$\tau  = \left( {10\hat i} \right) \times \left( {2\hat i + 3\hat j} \right)$

$\tau  = 30\hat k\;{\rm{N}} \cdot {\rm{m}}$

Thus, the torque acting on bar magnet is $30\hat k\;{\rm{N}} \cdot {\rm{m}}$.