Tag: magnetic field due to bar magnet

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Questions Related to magnetic field due to bar magnet

Multiple choice 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

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?

  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. $2\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}{\text{T}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
$\tau _{1}=\tau _{2}$
$MB _{1}sin\theta _1=MB2sin\theta _2$

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

Multiple choice 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

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

  1. $2\ Am^{2}$
  2. $4\ Am^{2}$
  3. $6\ Am^{2}$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

Multiple choice 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

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?

  1. $26$J $T^{-1}$
  2. $2.6$J $T^{-1}$
  3. $0.26$J $T^{-1}$
  4. $0.026$J $T^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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} $
Multiple choice 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

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:

  1. $30$ N m
  2. $30\sqrt{3}$ N m
  3. $60$ N m
  4. $60\sqrt{3}$ N m
Reveal answer Fill a bubble to check yourself
A Correct answer
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

Multiple choice 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

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?

  1. $0.075$N m
  2. $0.060$N m
  3. $0.081$N m
  4. $0.091$N m
Reveal answer Fill a bubble to check yourself
C Correct answer
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$

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

$=0.08125=0.08\ Nm$

Multiple choice 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

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 :

  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 $
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $30\hat k\;{\rm{N}} \cdot {\rm{m}}$.

Multiple choice 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

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

  1. $\dfrac { p } { \sqrt { 2 } \pi }$
  2. $\sqrt { 2 } p$
  3. $\dfrac { p } { 2 \pi }$
  4. $\dfrac { \sqrt { 2 } p } { \pi }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Pitch p = v*cos(theta)T = v*cos(theta)(2*pi*m)/(qB). Radius r = (mv*sin(theta))/(qB). Given theta = 45 degrees, sin(45) = cos(45) = 1/sqrt(2). Thus, p = (v*2*pi*m)/(sqrt(2)*qB) and r = (mv)/(sqrt(2)*qB). Comparing these, r = p/(2*pi).