Tag: magnetic effects of current and magnetism

Questions Related to magnetic effects of current and magnetism

If a solenoid is having magnetic moment of $0.65 JT^{-1}$ 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 $30^0$ with the direction of applied field is?

  1. $0.075$N m

  2. $0.080$ N m

  3. $0.081$ N m

  4. $0.091$ N m


Correct Option: C
Explanation:
Given that, $M=0.65 JT^{-1}$ , $B=0.25\ T$ and $\theta= 30^o$

$\therefore$ Torque on the solenoid $, \tau =MB\sin \theta =0.65\times 0.25\times \sin 30^o$
$=0.65\times 0.25\times \dfrac{1}{2}=0.08125=0.081\ Nm$

So, option $(C)$ is correct. 

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 $30^0$, will be then

  1. $30$Nm

  2. $30 \sqrt3$Nm

  3. $60$Nm

  4. $60 \sqrt3$Nm


Correct Option: A
Explanation:

Torque experienced by a magnet suspended in a uniform magnetic field B is given by
$T = MB sin \theta$
Here, $M = 200 A m^2, B = 0.30 N A^{-1}m^{-1}  and \  \theta = 30^0$
$\therefore T = 200 \times 0.30 \times sin 30^0$
$T = 30 N m$

A magnetic needle kept non parallel to the magnetic field in a non uniform magnetic field experience.

  1. A force but not a torque

  2. A torque but not a force

  3. Both a force and a torque

  4. Neither a force nor a torque


Correct Option: C
Explanation:

Magnetic needle can be considered as a magnetic dipole since it is placed in a non uniform magnetic field it will experience af force as magnetic force on both magnetic force will be different. Also, needle is placed non parallel to the field, so it will experience a torque also.

Therefore, C is correct option. 

A uniform horizontal magnetic field of $7.5\times 10^{-2}$T is set up at angle of $30^o$ with the axis of an solenoid and the magnetic moment associated with it is $1.28$J $T^{-1}$. Then the torque on it is?

  1. $4.8\times 10^{-2}$N m

  2. $1.6\times 10^{-2}$N m

  3. $1.2\times 10^{-2}$N m

  4. $4.8\times 10^{-4}$N m


Correct Option: A
Explanation:

Torque, $\tau =MB\sin\theta$
Here, $M=1.28J$ $T^{-1}$, $B=7.5\times 10^{-2}T, \theta =30^o$
$\therefore \tau =1.28\times 7.5\times 10^{-2}\sin 30^o$
$=1.28\times 7.5\times 10^{-2}\times \dfrac{1}{2}$
$=0.64\times 7.5\times 10^{-2}=4.8\times 10^{-2}$N m.