Tag: torque on current carrying loop

Questions Related to torque on current carrying loop

A flat coil carrying a current has a magnetic moment $\vec{\mu}$. It is placed in a magnetic field $\vec B$ such that $\vec{\mu}$ is antiparallel to $\vec B$. 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. not in equilibrium

  2. in stable equilibrium

  3. in unstable equilibrium

  4. in neutral equilibrium

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Correct Option: C
Explanation:

At  present situation force of magnetic moment = $ \vec{\mu} \times \vec{B} =0 $
If it is disturbed from this position, the force that would develop would take it away from this current position and destabilize the current equilibrium. Hence it is at untstable equilibrium.
In stable equilibrium, the force developed on the system upon disturbing brings the system back to its original position. Here the opposite is the case.
In neutral equilibrium, upon being disturbed from the original position, the system attains a new equilibrium at the new position.   

A dipole of dipole moment p is kept at the centre of a ring of radius R and charge Q. The dipole moment has direction along the axis of the ring. The resultant force on the ring due to the dipole 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. zero

  2. $\frac{k P Q}{R^3}$

  3. $\frac{2k P Q}{R^3}$

  4. $\frac{k P Q}{R^3}$ only if the charge is uniformly distributed on the ring

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Correct Option: A

A current loop in a 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. can be in equilibrium in one orientation.

  2. can be in equilibrium in two orientations, both the equilibrium states are unstable.

  3. can be in equilibrium in two orientations, one stable while the other is unstable.

  4. experiences a torque where the field is uniform or non-uniform in all orientations.

Show answer
Correct Option: C
Explanation:

A current loop is a magnetic field is in equilibrium in two orientations one is stable and another unstable.
$\because \vec{\tau} = \vec M \times \vec B = M  B  sin  \theta$
If $\theta = 0^o \Rightarrow \tau = 0$    (stable)
If $\theta = \pi \Rightarrow \tau = 0$     (unstable)
Do not experience a torque in some orientations
Hence option (c) is correct.

A current carrying loop lies on a smooth horizontal place. Then

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. it is possible to establish a uniform magnetic field in the region so that the loop starts rotating about its own axis.

  2. it is possible to establish a uniform magnetic field in the region so that the loop will tip over about any of the point.

  3. it is not possible that loop will tip over about any of the point whatever be the direction of established magnetic field (uniform).

  4. both A and B are correct.

Show answer
Correct Option: B
Explanation:

As the loop is placed in horizontal plane, so area vector is along vertical direction. From $\vec {\tau}=I(\vec A\times \vec B)$, as $\vec A$ is in vertical direction, $\vec {\tau}$ would be the plane of loop only. So, option (a) is wrong because for rotation of loop about its own axis $\vec {\tau}$ must be along vertical direction. (b) is correct because we can produce torque in the plane of the loop and due to this the loop can tip over.

The plane of a rectangular loop of wire with sides $0.05\ m$ and $0.08\ m$ is parallel to a uniform magnetic field of induction $1.5\times 10^{-2}\ T$. A current of  $10.0\ ampere$ flows through the loop. If the side of length $0.08\ m$ is normal and the side of length 0.05m is parallel to the lines of induction, then the torque 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. $6000\ Nm$

  2. $zero$

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

  4. $6\times 10^{-4}\ Nm$

Show answer
Correct Option: D
Explanation:

Torque $\tau$ acting on a current carrying coil of area A placed in a magnetic field of induction B is given by,


$\tau=NIBA \sin\theta$

where $I=$ current in the coil, $\theta=$ angle which the normal the plane of the coil makes with the lines of induction $B$.

Here, $N=1, B=1.5\times 10^{-2}\ T$

$A=0.05\times 0.08=40\times 10^{-4}\ m^2$


$I=10.0 amp, \theta=90^o=\pi /2$

$\tau=(1.5\times 10^{-2})(10.0)\times (1)(40\times 10^{-4}\sin (\pi /2)$

$\tau =6\times 10^{-4}\ Nm$

When a current carrying coil is placed in a uniform magnetic field with its magnetic moment anti-parallel to the 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. torque on it is maximum

  2. torque on it is zero

  3. potential energy is maximum

  4. dipole is in unstable equilibrium

Show answer
Correct Option: B,C,D
Explanation:

Given, $\vec{M} \nparallel \vec{B}$, It is anti parallel.
$\therefore \vec{M} \times \vec{B} =0$ since angle is $180^{\circ}$
Potential energy $U =-\vec{M}.\vec{B}= -MB \cos 180^{\circ}= MB \times 1= MB$
Potential energy is maximum is an equilibrium but is an unstable equilibrium.

A coil carrying electric current is placed in uniform magnetic field. Then : 

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. torque id formed

  2. emf is induced

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

  4. none of the above

Show answer
Correct Option: A
Explanation:

As magnetic field is constant therefore there will be no change in flux therefore no induced emf. current carrying coil has magnetic dipole moment. Hence a torque p = m x B acts on it in magnetic field.

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

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. inclined at 45$^o$ to the magnetic field

  2. inclined at any arbitrary angle to the magnetic field

  3. parallel to the magnetic field

  4. perpendicular to magnetic field

Show answer
Correct Option: D
Explanation:

$\tau = (\vec M \times \vec B)$, where $|\vec M| = i .A$

$= M B   sin  \theta$
where
$\theta$ is angle between Magnetic moment & $\vec B$.
For $\theta = 0   \vec{\tau} = 0$ & coil is in stable equilibrium.
Hence plane of coil must be perpendicular to magnetic field.

A current carrying loop is free to turn in a uniform magnetic field. The loop will then come into equilibrium when it's plane is inclined at

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. $0^{0}$ to the direction of the field.

  2. $45^{0}$ to the direction of the field.

  3. $90^{0}$ to the direction of the field.

  4. $60^{0}$to the direction of the field.

Show answer
Correct Option: C
Explanation:

loop will come to equilibrium when torque is zero
So, $\vec{\tau} = 0 = \vec{M}\times \vec{B}$
                            $= MB sin \theta$
                            $= NiA Bsin \theta$
So, angle between area vector and field is zero than only $\theta$ is 0 so that $sin \theta = 0$.
So, for area vector along field direction plane is perpendicular to the field.

A coil of area $100\ cm^{2}$ having $500$ turns carries a current of $1\ m A$. It is suspended in a uniform magnetic field of induction $10^{-3}Wb /m^{2}$ . It's plane makes an angle of $60^{o}$ with the lines of induction. The torque acting on 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. $250\times 10^{-8}Nm$

  2. $25\times 10^{-8}Nm$

  3. $2.5\times 10^{-8}Nm$

  4. $0.2\times 10^{-8}Nm$

Show answer
Correct Option: A
Explanation:

$J=\vec { M } \times \vec { B } $
$=ni\ \vec { A}\times   \vec { B } $
$=1\times { 10 }^{ -3 }\times 100\times { 10 }^{ -4 } \times  { 10 }^{ -3 }\times 500\times \sin { { 30 }^{ o  } } $
$=25\times { 10 }^{ -7 }Nm.$

Since. Plane makes an angle of $60^o$ with the lines of induction, it will makes an angle of $30^o$ with it's normal.