Tag: torque on current carrying loop

Questions Related to torque on current carrying loop

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

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  1. a, c are correct

  2. a, b, c are correct

  3. b,c,d are correct

  4. a,b,d are correct

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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

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  1.  $ir\bar{B} _{o}$

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

  3. $0$

  4. $\pi  ri\bar{B} _{o}$

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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

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  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

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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

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  1. $P _{m}$B

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

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

  4. $P _{m}$/B

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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

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  1. force only

  2. torque only

  3. both torque and force

  4. neither torque nor force

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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

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  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

Show answer
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.

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  1. 10

  2. 20

  3. 22

  4. 30

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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.

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  1. $\sqrt3 5X$

  2. $\sqrt3 X$

  3. $\sqrt3 2X$

  4. $\sqrt3 3X$

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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 $\theta =60$
                                        =${\sqrt3} X$

A circular coil of $25$ turns and radius of $12$cm is placed in a uniform magnetic field of $0.5$ T normal to the plane of coil. If the current in the coil is $5$A, then total torque experienced by the coil is

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  1. $1.5$N m

  2. $2.5$N m

  3. $3.5$ N m

  4. zero

Show answer
Correct Option: D
Explanation:

Here, n = $25$ turns, r = $12$cm, B = $0.5$T
Since the coil is placed in uniform magnetic field normal to the place of the coil. Hence the angle between magnetic moment and magnetic field direction is zero $(i.e. \theta = 0)$
$therefore = mB sin \theta = mB sin 0$
$\therefore T = 0$

A closely wound solenoid of 2000 turns and area of cross-section $1.5 \times 10^{-4}m^2$ carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{-2}T$, making an angle of 30$^o$ with the axis of the solenoid. The torque on the solenoid will be

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

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

  2. $1.5 \times 10^{-3} N -m$

  3. $1.5 \times 10^{-2} N -m$

  4. $3 \times 10^{-2} N -m$

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Correct Option: C
Explanation:

Given :   $N = 2000$ turns        $A = 1.5\times 10^{-4} m^2$              $I = 2.0$ A                $B = 5\times 10^{-2}$ T            $\theta = 30^o$

Torque    $\tau = NI (A\times B) =NIAB \sin \theta$                 
$\therefore$  $\tau = (2000) (2.0) (1.5\times 10^{-4}) (5\times 10^{-2}) (0.5) $                           $(\because \sin 30^o =0.5)$
$\implies$   $\tau = 1.5\times 10^{-2}$  $N-m$