Tag: force and torque on a current carrying rectangular loop in a uniform magnetic field

Questions Related to force and torque on a current carrying rectangular loop in a uniform magnetic field

The torgue required to keep a magnet of length $20cm$ at $30^o$ to a uniform field is $2 \times^{-5}N-m$. The magnetic force on each pole is 

  1. $2\times ^{-4}N$

  2. $4 \times 10^{-4}N$

  3. $1\times 10^{-3}N$

  4. $2\times 10^{-6}N$


Correct Option: A
Explanation:
We have,

Length of magnetic field is $0.20m$, Magnetic field $2\times10^{-5}Am$

So, $\tau=mB\sin30^0=0.20\times2\times10^{-5}\times\dfrac{1}{2}=2\times10^{-4}N$

A magnetic dipole placed in two perpendicular magnetic fields $\vec{B}$ and $\vec{B} _\circ{}$ is in equilibrium making an angle $\theta $ with $\vec{B}$ then

  1. $B = B _\circ{}$

  2. $B \cos \theta = B _\circ{}\sin\theta $

  3. $B \sin \theta = B _\circ{}\cos\theta $

  4. $B = B _\circ{} \tan \theta$


Correct Option: C
Explanation:

For equilibrium of dipole net torque on the loop should be zero.
So, $\vec{Z} = \vec{M}\times \vec{B}$
$\vec{Z} _1 = \vec{Z} _o$
$\vec{M}\times \vec{B} = \vec{M}\times \vec{B} _o$
$MB \sin \theta = M B _o \sin (90^o-\theta)$
$B \sin\theta = B _o \cos \theta$   $(\because \sin(90^o-\theta) = \cos \theta)$

A bar magnet of moment $\overline{M}$ is in a magnetic field of induction $\overline{B}$. Then the couple is:

  1. $\overline{M}\times \overline{B}$

  2. $\overline{B}\times \overline{M}$

  3. $\overline{M} . \overline{B}$

  4. $\overline{B} . \overline{M}$


Correct Option: A
Explanation:

$Torque = IAB \sin\theta$
$m = IA$
$T = mB \sin\theta$
$T = m \times B$

A current carrying a coil suspended freely in a uniform magnetic field is in stable equilibrium, if the angle between its magnetic dipole moment vector and the magnetic field is 

  1. $180^0$

  2. zero

  3. $45^0$

  4. $90^0$


Correct Option: B
Explanation:

A current carrying coil behaves as a magnetic dipole. 

Now we know that most stable position for dipole placed in a magnetic field is when angle between its magnetic moment and the external magnetic field is zero. 
In this position potential energy of the system is minimum therefore, it is in stable equilibrium position.

A current carrying loop is placed in a uniform magnetic field. The torque acting on it does depend upon

  1. shape of the loop keeping perimeter fix

  2. area of the loop

  3. value of the current

  4. magnetic field


Correct Option: A,B,C,D
Explanation:

Torque $\vec{\tau} = \vec{\mu} \times \vec{B}$ where $\vec{\mu} = iA\hat{n}$
$i$ is the current in the loop and $A$ is the area of the loop. $\hat{n}$ is a unit vector perpendicular to the plane of the loop.
Area will depend on the shape. Hence, $\vec{\tau}$ will depend on  area, current and  magnetic field.

A coil carrying electric current is placed in a uniform magnetic field with its axis making a nonzero angle $\theta$ with the field, then

  1. torque is applied on the coil

  2. emf is induced

  3. both (a) and (b) are correct

  4. a net force acts on the coil


Correct Option: A
Explanation:

Due to rotating current in the coil, a magnetic moment $M$ is produced. 
The torque exerted on the coil=$\vec{M}\times \vec{B}=MB\sin\theta\neq 0$ for $\theta\neq 0$.

A current carrying coil is subjected to a uniform magnetic field. The coil will orient so that its plane becomes

  1. inclined at $45^{o}$ to the magnetic field

  2. inclined at an arbitrary angle to the field

  3. parallel to the magnetic field

  4. perpendicular to the magnetic field


Correct Option: D
Explanation:

A current carrying coil behaves as a magnetic dipole. Therefore, in a uniform magnetic field coil will get aligned such that the dipole moment of the coil is parallel to the magnetic field. And we know that dipole moment of a coil is perpendicular to its plane.
Therefore, coil will align itself such that its plane is perpendicular the direction of magnetic field.
option(D)

A magnetic field is produced and directed along y-axis. A magnet is placed along x-axis .The direction of the torque on the magnet is:

  1. in the x-y plane

  2. along z-axis

  3. along y-axis

  4. torque will be zero


Correct Option: B
Explanation:

$\tau =\vec{M}\times \vec{B}$
$\hat{i}\times \hat{j}=\hat{k}$
The direction of torque will be always $Z-axis (+ve)$

There is no couple acting when two bar magnets are placed co-axially separated by a distance because :

  1. there are no forces on the poles

  2. the forces are parallel and their lines of action do not cincide

  3. the forces are perpendicular to each other

  4. the forces act along the same line


Correct Option: D
Explanation:

$\vec{\tau} =\vec{M}\times \vec{B}=0$

$MB\sin \theta =0$
$\sin \theta =0$
$\Rightarrow \theta =0$
$\Rightarrow $ Magnetic field lines of both the magnets are in the same line which results in forces due to both magnets act along the same line

A circular loop of area 0.02 m$^2$ carrying a current of 10A, is held with its plane perpendicular to a magnetic field induction 0.2 T. The torque acting on the loop is

  1. 0.01 Nm

  2. 0.001 Nm

  3. zero

  4. 0.8 Nm


Correct Option: C