Tag: magnetic field due to bar magnet

Questions Related to magnetic field due to bar magnet

A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is $60^o$, and one of the fields has a magnitude of $1.2\times 10^{-2} T$. If the dipole comes to stable equilibrium at an angle of $15^o$ with this field, what is the magnitude of the other field?

physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. $3\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}{\text{T}}$

  2. $\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}{\text{T}}$

  3. $6\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}{\text{T}}$

  4. $2\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}{\text{T}}$

Show answer
Correct Option: A
Explanation:
Here, $\theta=60^{0}, B1=1.2\times 10^{−2}tesla,$
$\theta _1=15^{0}, \theta _{2}=60^{0}−15^{0}=45^{0}.$
In equilibrium, torque due to two fields must balance i.e.
$\tau _{1}=\tau _{2}$
$MB _{1}sin\theta _1=MB2sin\theta _2$

$\implies 1.2\times 10^{-2}\times sin15^{\circ}=B _2sin(60-15)^{\circ}=B _2sin45^{\circ}$
$\implies B _2=4.4\times 10^{-3}T$

A short bar magnet experiences a torque of magnitude $0.64\ J$. When it is placed in a uniform magnetic field of $0.32\ T$, making an angle of $30^{\circ}$ with the direction of the field. The magnetic moment of the magnet is

physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. $2\ Am^{2}$

  2. $4\ Am^{2}$

  3. $6\ Am^{2}$

  4. None of these

Show answer
Correct Option: B
Explanation:

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

A short bar magnet placed with its axis at $30^o$ with a uniform external magnetic field of $0.35$ T experiences a torque of magnitude equal to $4.5\times 10^{-2}$N m. The magnitude of magnetic moment of the given magnet is?

physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. $26$J $T^{-1}$

  2. $2.6$J $T^{-1}$

  3. $0.26$J $T^{-1}$

  4. $0.026$J $T^{-1}$

Show answer
Correct Option: C
Explanation:
Given:
The angle made by the magnetic field is, $θ=30^o$
The magnetic field in the region is, $B=0.35\ T$
The torque acting on the bar magnet is $\tau=4.5\times 10^{-2}\ Nm$

The torque acting on the magnet is given by:

$\tau=MB\ sin\ θ$

$ 4.5\times 10^{-2}= M\times 0.35 \times (sin 30^o)$

$⟹M=0.26\ JT^{-1} $

A bar magnet has a magnetic moment of $200$ A $m^2$. The magnet is suspended in a magnetic field of $0.30$N $A^{-1}m^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^o$, will be:

physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. $30$ N m

  2. $30\sqrt{3}$ N m

  3. $60$ N m

  4. $60\sqrt{3}$ N m

Show answer
Correct Option: A
Explanation:

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

If a solenoid is having magnetic moment of $0.65$J $T^{-1}$ is free to turn about the vertical direction and has a uniform horizontal magnetic field of $0.25$T applied. What is the magnitude of the torque on the solenoid when its axis makes an angle of $30^o$ with the direction of applied field?

physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. $0.075$N m

  2. $0.060$N m

  3. $0.081$N m

  4. $0.091$N m

Show answer
Correct Option: C
Explanation:

Given:

The magnetic moment of the solenoid is $M=0.65\ JT^{-1}$
The magnetic field applied across the solenoid is $\ B=0.25\ T$
The angle made by the axis of the solenoid is $\theta=30^o$

The torque experienced by the solenoid is given by:
$\therefore \tau =MB\sin \theta$

$ =0.65\times 0.25\times \sin 30^o$

$=0.65\times 0.25\times \dfrac{1}{2}$

$=0.08125=0.08\ Nm$

A magnet of magnetic moment $10 \hat{i} A-m^2 $ is placed along the x-axis in a magnetic field $ \overline{B} = ( 2 \hat {i} + 3 \hat{j} )  T $ . The torque acting on bar magnet is :

physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. $ 20 \hat{i} + 30 \hat{k} N-m $

  2. $20 \hat{k}  N-m $

  3. $ 30 \hat{k} N-m $

  4. $ 20 \hat{i} + 30 \hat{j}  N-m $

Show answer
Correct Option: C
Explanation:

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

A charged particle enters a uniform magnetic field with velocity vector at an angle of $45 ^o$ with the magnetic field. The pitch of the helical path followed by the particle is $p.$ The radius of the helix will be

physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. $\dfrac { p } { \sqrt { 2 } \pi }$

  2. $\sqrt { 2 } p$

  3. $\dfrac { p } { 2 \pi }$

  4. $\dfrac { \sqrt { 2 } p } { \pi }$

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Correct Option: C