Tag: physics

Questions Related to physics

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

If a current carrying loop is placed in non uniform magnetic field, then the loop
a) experiences a force
b) experiences a torque
c) will develop induced current
d) oscillates

  1. a, c are correct

  2. a, b, c are correct

  3. b,c,d are correct

  4. a,b,d are correct


Correct Option: B
Explanation:

Current carrying loop in a non-uniform magnetic field experiences a torque, force and current is induced in the loop.

A conducting circular loop of radius $r$ carries a constant current $i$. It is placed in a uniform magnetic field $\bar{B} _{o}$ such that $\bar{B} _{o}$ is perpendicular to the plane of the loop. The magnetic force acting on the loop is

  1.  $ir\bar{B} _{o}$

  2. $2\pi ri \bar{B} _{o}$

  3. $0$

  4. $\pi  ri\bar{B} _{o}$


Correct Option: C
Explanation:

The direction of magnetic force acting on a small current carrying element is given by $\vec{dF}=i(\vec{dl}\times \vec{B})$

$\implies  \vec{F}=\oint i(\vec{dl}\times \vec{B})$ 
Since the vectors $\vec{dl}$ form a closed loop with angle between $\vec{dl}$ and $\vec{B}$ equal to $90^{\circ}$, the net force is zero.

A rectangular coil of wire carrying a current is kept in a uniform magnetic field. The torque acting on the coil will be maximum when

  1. the plane of the coil is perpendicular to the field

  2. the normal to the plane of the coil is parallel to the field

  3. the normal to the plane of the coil is perpendicular to the field

  4. the plane of the coil is making an angle of $45^o$ with the field


Correct Option: C
Explanation:

$\vec{\tau} = \vec{M}\times \vec{B}$
$= niAB \sin\theta$
So, $\sin \theta$ is maximum when $\theta = 90^o$
and $\theta$ is normal between normal of plane of coil and field.
So, the normal to the plane of the coil should be perpendicular to the field to produce maximum torque.

The torque acting on a magnetic dipole of moment $P _{m}$ when placed in a magnetic field is

  1. $P _{m}$B

  2. $\bar{P _{m}}\times \bar{B}$

  3. $\bar{P _{m}}.\bar{B}$

  4. $P _{m}$/B


Correct Option: B
Explanation:

Torque $(\tau) = \vec{M}\times \vec{B}$
                   $= \vec{P _m}\times \vec{B}$

A current carrying loop in a uniform magnetic field will experience

  1. force only

  2. torque only

  3. both torque and force

  4. neither torque nor force


Correct Option: B
Explanation:

A current carrying loop in a uniform magnetic field will experience torque only given by
$\vec{\tau}= \vec{M}\times \vec{B}$
Where M is magnetic moment of loop B is magnetic field.
Force is zero on a current carrying loop in a uniform magnetic field.
Note : Magnetic field must be uniform for net force to be zero.
For a non uniform field, net force may not be zero

A rectangular coil of wire carrying a current is kept in a uniform magnetic field. The torque acting on the coil will be zero when

  1. the plane of the coil is perpendicular to the field

  2. the normal to the plane of the coil is making an angle of 45$^o$ with the field

  3. the normal to the plane of the coil is perpendicular to the field

  4. the plane of the coil is parallel to the field


Correct Option: A
Explanation:

$\vec{\tau} = \vec{M}\times \vec{B}$
$= niAB \sin\theta$
So, for torque to be zero $\sin \theta$ should be zero and for that $\theta$ should be zero, where $\theta$ is angle between normal to the plane of coil and field.
So the plane of the coil and field should be perpendicular to each other

The coil of a galvanometer has $500$ turns and each turn has an average area of $3\times 10^{-4} m^2$. when a current of $0.5$ A passes through it. If a torque of $1.5 Nm$ is required for this coil carrying same current to set it parallel to a magnetic field, calculate the strength of the magnetic field.

  1. 10

  2. 20

  3. 22

  4. 30


Correct Option: B

If X amount of work is required to rotate a bar magnet in a magnetic field by $60^0$, from a position parallel to the field, What is the torque required to maintain it in new position.

  1. $\sqrt3 5X$

  2. $\sqrt3 X$

  3. $\sqrt3 2X$

  4. $\sqrt3 3X$


Correct Option: B
Explanation:

work done in rotating a magnet from $0$ to $60$ will be,

                        $X =$$M\times B\times (1- cos\theta)$
                            =$M\times B\times (0.5)$
                         $2X=M\times B$              taking $\theta =60$
      torqur required to maintain in that position will be ,
                               $\tau=M\times B\times Sin\theta$
                               $\tau=2X\times \dfrac{\sqrt(3)}{2}$     taking $\theta =60$
                                        =${\sqrt3} X$