Torque on current carrying loop - class-XII

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

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

A current carrying coil behaves as a magnetic dipole. 

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.

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 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)

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

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

$MB\sin \theta =0$
$\sin \theta =0$
$\Rightarrow \theta =0$
$\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

The torque on a current loop is given by tau = M x B, which has a magnitude of MB sin(theta), where theta is the angle between the area vector and the magnetic field. Since the plane of the loop is perpendicular to the magnetic field, the area vector is parallel to the field, making theta = 0 degrees. Thus, sin(0) = 0, resulting in zero torque.

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

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

$\vec{\tau} = \vec{M}\times \vec{B}$
$= niAB \sin\theta$
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.

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

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

$\vec{\tau} = \vec{M}\times \vec{B}$
$= niAB \sin\theta$
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 torque on a coil is given by tau = N I A B sin(theta). Here, the coil is parallel to the field, so theta = 90 degrees and sin(90) = 1. Plugging in the values: 1.5 = 500 * 0.5 * (3 * 10^-4) * B. Solving for B gives 1.5 = 0.075 * B, so B = 1.5 / 0.075 = 20 T.

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

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 $(i.e. \theta = 0)$
$therefore = mB sin \theta = mB sin 0$
$\therefore T = 0$

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

$\tau = MB sin\theta$
$4.5 \times 10^{-4}= M\times 0.5 \times 10^{-4}. sin 90^0$
$M= \dfrac{4.5 \times 10^{-4}}{0.5\times 10^{-4}}= 9 J/T$

The torque acting on a coil is given by,

Torque experienced by bar magnet is $\tau =MB\sin \theta $ where $M$magnetic moment, $B$ is magnetic field induction and $\theta $ is the angle between $M\,\,and\,\,B$

It is given that

  $ \tau =2.5\times {{10}^{-6}}N $

 $ B=5\times {{10}^{-5}}T $

 $ L=0.05m $

 $ 2.5\times {{10}^{-6}}=M\times 5\times {{10}^{-5}}\,\times \dfrac{1}{2} $

 $ M={{10}^{-1}} $

 $ M=m\times L $

 $ {{10}^{-1}}=m\times 0.05 $

 $ m=2 Am $

We know that,

$ \tau =MB\sin \theta  $

$ \tau \propto \sin \theta  $

So,

$ \dfrac{{{\tau } _{1}}}{{{\tau } _{2}}}=\dfrac{MB\sin {{\theta } _{1}}}{MB\sin {{\theta } _{2}}} $

$ \dfrac{{{\tau } _{1}}}{{{\tau } _{2}}}=\dfrac{\sin {{\theta } _{1}}}{\sin {{\theta } _{2}}} $

$ \dfrac{\tau }{\frac{\tau }{2}}=\dfrac{\sin {{90}^{0}}}{\sin {{\theta } _{2}}} $

$ \sin {{\theta } _{2}}=\dfrac{1}{2} $

$ {{\theta } _{2}}={{30}^{0}} $

Hence, the angle is ${{30}^{0}}$

Since,  $T _B=MB\sin\theta$ where $\theta$ is angle between body and the magnetic field$\implies $ shape of loop does not affect torque since $B$ depends on the area and current.

If the net force acting on the loop is zero then, 

Work done to rotate a magnet by 180 degrees is W = MB(cos(0) - cos(180)) = MB(1 - (-1)) = 2MB. W = 2 * 1.5 * 0.22 = 0.66 J. In the opposite position (180 degrees), the torque tau = MB sin(180) = 0.

On applying a uniform magnetic field on a current-carrying coil, the lines of force are at right angle to the plane of coil. Hence, the coil rotates in such a way that its plane becomes perpendicular to magnetic field.

The torque on a magnetic dipole (current loop) in a magnetic field is defined by the vector cross product of the magnetic moment and the magnetic field: tau = M x B.

Magnetic moment = $ I \vec A = 0.01 \times 10   A/m^2$ perpendicular to the field.
Maximum torque on the magnetic moment is when angle between magnetic moment and the field is $90^{\circ}= (I \vec A ) \times \vec B = I A B = 10 \times 0.01 \times 0.1 = 0.01 Nm $

The magnetic moment is defined as a vector relating the aligning torque on the object from an externally applied magnetic field to the field vector itself. The relationship is given by:

$ \tau = \vec{\mu} \times \vec{B} $

where  $\tau$ is the torque acting on the dipole and $B$ is the external magnetic field, and $\mu$  is the magnetic moment. Direction of torque is given by the right hand rule.

A current carrying loop behaves as a magnetic dipole. and we know that a dipole placed in uniform magnetic field only experiences torque.

Original moment M1 = I * L^2. After folding, the two halves are perpendicular. The new moment M2 is the vector sum of the moments of the two halves. Each half has moment M/2. The resultant is sqrt((M/2)^2 + (M/2)^2) = M/sqrt(2). The ratio M1/M2 = L^2 / (L^2/sqrt(2)) = sqrt(2).

Torque on the loop $\vec{\tau} = \vec{\mu} \times \vec{B}$ where $\vec{\mu}=i\vec{A}$ is the magnetic moment of the loop and $\vec{B}$  is the magnetic field.
$ \therefore \tau = i\vec{A} \times \vec{B}$
Since $\vec{B}$ is in the plane of the loop, $\vec{A} \perp \vec{B}$
$ \therefore \vec{\tau} = 10 \times ( 0.05 \times 0.08) \times( 1.5 \times 10^{-2})= 6 \times 10^{-4}N-m$

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 at the center of a ring produces a field that is symmetric. The force on a charge element dq of the ring is dF = dq * E_dipole. Due to the symmetry of the dipole field and the ring, the net force integrates to zero.

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.

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.

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

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.

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.

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

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.

$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.

Magnetic moment vector is parallel to that of field. Hence, Torque $=\vec { M } \times \vec { B } =Q\quad $

When current carrying coil is placed in the magnetic field torque acts on the coil. The torque acting on the coil will be zero if the coil and the magnetic field are perpendicular. Thus it orients such that the plane of coil is perpendicular to the field direction.

$A=400\times { 10 }^{ -4 }{ m }^{ 2 }$
 $n = 500$
$B=4\times { 10 }^{ -3 }$
Area vector makes complementary angle with field therefor, angle between magnetic field and normal of coil is $\theta=90^o-60^{ o }={30  }^{ o  }$
$i = 0.2$

$J= ni AB\sin\theta $
maximum at ,
$ \sin\theta=1$
$\therefore J=niAB=100\times 2\times 20\times 10\times 20\times { 10 }^{ -4 }=80 Nm$

Torque on magnetic dipole, $\hat i (A\times B)$
where $i$ is current 
$A$ is area vector normal to the area surface
$B$ is magnetic field

${ \tau } _{ 1 }=\vec { M } \times \vec { B }  = MB  \sin{ 60 }^{ o  }$
${ \tau } _{ 2 }=\vec { M } \times  \vec { B }  = MB  \sin{ 30 }^{ o }$

If coil is making an angle of $\theta$ with field, then it makes an angle of $(90 -\theta)$ with area vector, i.e. magnetic moment vector.

    $\therefore \quad \dfrac { { \tau } _{ 1 } }{ { \tau } _{ 2 } } =\dfrac { \sqrt { 3 }  }{ 1 } $

If plane is perpendicular to the field, normal vector will be parallel to field 
$\sin \theta=0$
$J = MB \sin\theta = 0$

The current flowing clockwise in an equilateral triangle has a magnetic field in the direction of $\hat { k }$.
$\tau =BiNA\sin { \theta  } $
$\tau =BiNA\sin { { 90 }^{ o } } $
$\tau =Bi\times \dfrac { \sqrt { 3 }  }{ 4 } { I }^{ 2 }\times 1$
[$\because $ Area of equilateral triangle $=\dfrac { \sqrt { 3 }  }{ 4 } { I }^{ 2 }$ and $N=1$]
(assuming)
$\Rightarrow { I }^{ 2 }=\dfrac { 4\tau  }{ \sqrt { 3 } Bi } \Rightarrow I=2{ \left[ \dfrac { \tau  }{ Bi\sqrt { 3 }  }  \right]  }^{ { 1 }/{ 2 } }$

Let us take an element at an angle $\theta$ subtending an angle $d \theta$

Here $E=2\times { 10 }^{ 5 }N{ C }^{ -1 },l=2cm,\tau =4Nm\quad $
Torque $\vec { \tau  } =\vec { p } \times \vec { E } =pE\sin { \theta  } $
$\therefore 4=p\times 2\times { 10 }^{ 5 }\times \sin { { 30 }^{ o } } $
or $p=4\times { 10 }^{ -5 }Cm$
$\therefore$ Charge $q=\cfrac { p }{ l } =\cfrac { 4\times { 10 }^{ -5 }Cm }{ 0.02m } =2\times { 10 }^{ -3 }C=2mC$

Potential energy U = -M * B = -MB cos(theta). Stable equilibrium occurs at the minimum potential energy, which is when cos(theta) = 1, meaning theta = 0 degrees.

The torque acting on the coil $\tau$=|$\bar m$ X $\bar B$ | =$ mB\sin\Theta$
Here the circular coil is placed normal to the direction of magnetic field then the angle between the. direction of magnetic moment ($\bar m$) and magnetic field ($\bar B$) is zero, then
$\tau$ = $mBsin\Theta$ = $mBsin 0$ = 0                $\tau = 0$

Given that,

Given that,

Torque, $\bar{T} = \bar{m} \times \bar{B} = mB sin \theta$
and
magnetic potential energy $U _m = \bar{m}. \bar{B} = mB cos \theta$ at
$\theta = 0^0$ the dipole will be in most stable position
$T = mB sin\theta = mB sin 0^0 = 0$
and $U _m = -mB cos\theta = -mB cos 0^0 = -mB$

Here, $B = 900 Gauss = 900 \times 10^{-4}T. = 9 \times 10^{-2}T$
$T = 0.02 N m and \theta = 30^0$
$\therefore T = mB sin \theta$
$\Rightarrow 0.02 = m \times 9 \times 10^{-2} \times sin 30^0$
$m = \dfrac{0.02 \times 2 }{9 \times 10^{-2}}= 0.44 A m^2$.

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

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.

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.