Informative line

Magnetic Moment And Torque

Practice to calculate the magnetic moment of the coil and on a square loop placed in uniform magnetic field, and Find the torque experienced by dipole in uniform magnetic field and magnetic dipole having dipole moment.

Magnetic Moment on a Square Loop Placed in Uniform Magnetic Field 

  • Consider a square loop of side 'a' carrying current \(\text I\), placed in a uniform magnetic field as shown in figure.  

  • Force on side 1 due to magnetic field \(\vec B\) 

Since,  length (\(\vec L\)) and magnetic field \((\vec B)\) are perpendicular to each other. So, angle between \(\vec L\) and \(\vec B\)  is \(\theta = \) 90º

Force on side 1

 \(\vec F _1 = I (\vec L × \vec B)\) 

\(= ILB \,sin\,\theta\)

\(I \,a \, B \,sin \,90º \)

where,  \(\theta = \)  90º, L = a

\(I a \,B \)       (into the page)

 

  • Force on side 2 due to magnetic field \(\vec B\) 

Since, length (\(\vec L\)) and magnetic field \((\vec B)\) are parallel to each other. So, angle between \(\vec L\) and \(\vec B\)  is \(\theta = \) 0º

Force on side 2 

  \(\vec F _2 = I (\vec L × \vec B)\) 

         \(= ILB \,sin\,\theta\)

          = \(I \,a \, B \,sin \,0º \) 

(where, \(\theta = \)  0º, L = a)

          = 0                         

  • Force on side 3 due to magnetic field \(\vec B\) 

Since,  length (\(\vec L\)) and magnetic field \((\vec B)\) are perpendicular to each other. So, angle between \(\vec L\) and \(\vec B\)  is \(\theta = \) 90º

Force on side 3

 \(\vec F _3 = I (\vec L × \vec B)\) 

 \(\vec F_3= ILB \,sin\,\theta\)

\(\vec F_3 =I \,a \, B \,sin\, 90º \)

where, \(\theta = \)  90º, L = a

\(\vec F_3 = I\,a\,B \)   (outside the page)

  • Force on side 4 due to magnetic field \(\vec B\) 

Since,  length (\(\vec L\)) and magnetic field \((\vec B)\) are parallel  to each other. So, angle between \(\vec L\) and \(\vec B\)  is \(\theta = \) 0º

Force on side 4

 \(\vec F _4 = I (\vec L × \vec B)\) 

  \(\vec F_4 = ILB \,sin\,\theta\)

 \(\vec F_4 =I \,a \, B \,sin\, 0º \)

 where, \(\theta = 0°\), L = a

\(\vec F_4 = 0\)                                        

  • The figure represents edge view of loop slighting down sides 1 and 3 shows that the magnetic force \(\vec F_1\) and \(\vec F_3 \)  exerted on these sides creates a torque that tend to twist the loop.
  • The cross in the left circle represent current in wire 1 moving away from you and dot in right side in a wire represent current in wire 3 coming towards you.
  • Hence, we can say that two forces are acting on loop in opposite directions. So, the loop will rotate. 

  • Thus, the magnitude of Torque

\(\tau = F_1 × \dfrac{a}{2} + F_3 × \dfrac{a}{2} \)

(\(\because\)when net force on a body in zero, torque about every point is same)

\(\tau = IaB \dfrac{a}{2} + IaB \dfrac{a}{2} \)

       \(\tau = I a^2 B \)

        \(\tau = I AB\)  , where A = Area of square plate = a2

  • If loop is not parallel to the magnetic field 

              Torque \(\vec \tau = IAB \,sin \,\theta\)

             \(\vec \tau = I(\vec A × \vec B)\)      

where, \(\vec A\)  is the area vector.

Note :  Using right hand thumb rule , curl fingers in the direction  of current, then thumb gives the direction of \(\vec A\) which is inside the page.

  • The term \(I \vec A\) is defined as the magnetic dipole moment and denoted by  \(\vec \mu\).

              \( \vec \mu=I \vec A\)

Then, torque \(\vec \tau = \vec \mu × \vec B\)

Unit:

Unit  of magnetic moment \((\vec \mu) = Ampere × m^2\)

  • If a coil of wire contains N loop of same area 

            \(\vec \mu = N I \, \vec A\)

Illustration Questions

A rectangular coil of area A = 2 m × 4 m, carrying current  \(\text I = 5\,mA\)  consists of N =100 turns of wire. If this coil is placed in magnetic field, calculate magnetic moment of the coil.

A 8 Am2

B 6 Am2

C 3 Am2

D 4 Am2

×

Magnetic moment ( \(\mu\)) is given as 

\(\mu = N\,I\,A\)

where,

N = Number of turns 

\(\text I = \) Current in coil

A = Area

Given : \(N = 100\)\(\text I = 5\,mA\)\(A = 2\,m × 4\, m\)

\(\mu = NIA\)

\(\mu = 100× 5 × 10^{-3} × 2 × 4\)

\(\mu = 4 \,\text A m^2\)

A rectangular coil of area A = 2 m × 4 m, carrying current  \(\text I = 5\,mA\)  consists of N =100 turns of wire. If this coil is placed in magnetic field, calculate magnetic moment of the coil.

A

8 Am2

.

B

6 Am2

C

3 Am2

D

4 Am2

Option D is Correct

Magnetic Moment of a Charged Particle Moving in Circular Motion with Constant Angular Velocity

  • Consider a charge q, moving in a circular path with constant angular velocity  \(\omega \). The radius of circular path is r.

  • Magnetic moment is given as, 

           \(\mu = NIA\)

        when only one turn is considered N =1

         \(\mu = IA\)

         \(\mu = I× \pi r^2\)        [\(A = \pi r^2\) ]     ............(1)

  • Current (\(I\)) is given as: 

\(I = \dfrac{q}{T}\)       

where, q = charge,T = Time

\(I = \dfrac{q}{\dfrac{2\pi}{\omega}}\)                   \(\left[\therefore T = \dfrac{2\pi}{\omega}\right]\)

So,     \(I = \dfrac{\omega q }{2\pi}\)     ............(2)

  • Thus, magnetic moment is = \(IA\)

\(\mu = \dfrac{\omega q}{2\pi} × \pi\, r^2\)                    \(\left[I = \dfrac{\omega q}{2\pi}\right]\)

\(\mu = \dfrac{\omega\, q\, r^2}{2} \)

Illustration Questions

Calculate the magnetic moment of a charged particle having charge \(q = 2 \,\mu C\), moves in a circular path of radius r= 1m with constant angular velocity \(\omega = \) 3 rad/sec.

A  \(12\, \mu A\,m^2\)

B \(3\, \mu A\,m^2\)

C \(6\, \mu A\,m^2\)

D \(8\, \mu A\,m^2\)

×

A charge q, moving in a circular path with constant angular velocity  \(\omega \) . The radius of circular path is r.

image

Magnetic moment is given as, 

\(\mu = NIA\)

        when only one turn is considered N =1

\(\mu = IA\)

\(\mu = I× \pi r^2\)        [\(A = \pi r^2\) ]     ............(1)

\(\mu=\pi\,\text I(1)^2\)                         [ r=1m ]

\(\mu=\pi\,\text {I Am}^2\)

 

Current (\(I\)) is given as: 

\(I = \dfrac{q}{T}\)       

where, q = charge,T = Time

\(I = \dfrac{q}{\dfrac{2\pi}{\omega}}\)                   \(\left[\therefore T = \dfrac{2\pi}{\omega}\right]\)

So,     \(I = \dfrac{\omega q }{2\pi}\)     ............(2)

  \(I = \dfrac{3× 2× 10^{-6}}{2\pi}\)

  \(I = \dfrac{3\mu}{\pi} A\)

Thus, magnetic moment is = \(IA\)

\(\mu = \pi × \dfrac{3\mu}{\pi} A m^2\)

\(\mu = 3\,\mu A\,m^2\)

Calculate the magnetic moment of a charged particle having charge \(q = 2 \,\mu C\), moves in a circular path of radius r= 1m with constant angular velocity \(\omega = \) 3 rad/sec.

A

 \(12\, \mu A\,m^2\)

.

B

\(3\, \mu A\,m^2\)

C

\(6\, \mu A\,m^2\)

D

\(8\, \mu A\,m^2\)

Option B is Correct

Torque Experienced by Dipole in Uniform Magnetic Field

  • If loop is not parallel to the magnetic field. 

  • Torque \(\vec \tau = IAB \,sin\, \theta\)

             \(\vec \tau = I(\vec A × \vec B)\)          where \(\vec A\)  is the area vector

Note :  Using right hand thumb rule, curl fingers in direction of current, then thumb gives the direction of \(\vec A\) which is inside the page.

  • The term \(I \vec A\) is defined as the magnetic dipole moment and denoted by  \(\vec \mu\).

 \( \vec \mu=I \vec A\)

Then, torque \(\vec \tau = \vec \mu × \vec B\)

Unit:

Unit  of magnetic moment, \((\vec \mu) = Ampere × m^2\)

  • If a coil of wire contains N loop of same area 

            \(\vec \mu = N I \, \vec A\)

  • It is called as magnetic dipole moment.

Illustration Questions

Find the torque experienced by magnetic dipole having dipole moment \(\mu = \) 5 Am2, placed in a magnetic field B = 1 T, as shown in figure. Given, \(\theta=\)  37º  [sin 37º = 3/5]

A 8 N-m

B 4 N-m

C 6 N-m

D 3 N-m

×

Torque experienced by dipole is given as 

 \(\vec \tau=\vec \mu× \vec B\)

\(\vec \tau = \mu B \, sin \,\theta\)

where, \(\mu\) is dipole moment

B is magnetic field  

Given : \(\mu = \) 5 Am2,  B = 1 T, \(\theta = \) 37º

\(\tau = 5× 1 × sin\, 37 º\)

\(\tau = 5× 1 × \dfrac{3}{5}\)

\(\tau = 3 \,\text{N-m}\)

Find the torque experienced by magnetic dipole having dipole moment \(\mu = \) 5 Am2, placed in a magnetic field B = 1 T, as shown in figure. Given, \(\theta=\)  37º  [sin 37º = 3/5]

image
A

8 N-m

.

B

4 N-m

C

6 N-m

D

3 N-m

Option D is Correct

Magnetic Moment on a Square Loop placed in Uniform Magnetic Field 

  • Consider a square loop of side 'a' carrying current \(\text I\), placed in a uniform magnetic field as shown in figure.  

  • Force on side 1 due to magnetic field \(\vec B\) 

Since,  length (\(\vec L\)) and magnetic field \((\vec B)\) are perpendicular to each other. So, angle between \(\vec L\) and \(\vec B\)  is \(\theta = \) 90º

Force on side 1

 \(\vec F _1 = I (\vec L × \vec B)\) 

\( ILB \,sin\,\theta\)

\(I \,a \, B \,sin \,90º \)

where,  \(\theta = \)  90º , L = a

\(I a \,B \)       (into the page)

  • Force on side 2 due to magnetic field \(\vec B\) 

Since,  length (\(\vec L\)) and magnetic field \((\vec B)\) are parallel to each other. So, angle between \(\vec L\) and \(\vec B\)  is \(\theta = \) 0º

Force on side 2 

  \(\vec F _2 = I (\vec L × \vec B)\) 

         \(= ILB \,sin\,\theta\)

          = \(I \,a \, B \,sin \,0º \) 

(where, \(\theta \)=  0º, L = a)

          = 0                         

  • Force on side 3 due to magnetic field \(\vec B\) 

Since,  length (\(\vec L\)) and magnetic field \((\vec B)\) are perpendicular to each other. So, angle between \(\vec L\) and \(\vec B\)  is \(\theta = \) 90º

Force on side 3

 \(\vec F _3 = I (\vec L × \vec B)\) 

 \(\vec F_3= ILB \,sin\,\theta\)

\(\vec F_3 =I \,a \, B \,sin\, 90º \)

where, \(\theta = \)  90º, L = a

\(\vec F_3 = I\,a\,B \)   (outside the page)

  • Force on side 4 due to magnetic field \(\vec B\) 

Since,  length (\(\vec L\)) and magnetic field \((\vec B)\) are parallel  to each other. So, angle between \(\vec L\) and \(\vec B\)  is \(\theta = \) 0º

Force on side 4

 \(\vec F _4 = I (\vec L × \vec B)\) 

  \(\vec F_4 = ILB \,sin\,\theta\)

 \(\vec F_4 =I \,a \, B \,sin\, 0º \)

 where, \(\theta = \)  0º, L = a

\(\vec F_4 = 0\)                                        

  • The given figure represents edge view of loop slighting down sides 1 and 3 shows that the magnetic force \(\vec F_1\) and \(\vec F_3\) exerted on these sides creates a torque that tend to twist the loop.
  • The cross in the left circle represent current in wire 1 moving away from you and dot in right side in a wire represent current in wire 3 coming towards you.
  • Hence, we can say that two forces are acting on loop in opposite directions. So, the loop will rotate. 

Thus, the magnitude of Torque

\(\tau = F_1 × \dfrac{a}{2} + F_3 × \dfrac{a}{2} \)

(\(\because\)When net force on a body in zero, torque about every point is same.)

\(\tau = IaB \dfrac{a}{2} + IaB \dfrac{a}{2} \)

       \(\tau = I a^2 B \)

        \(\tau = I AB\) 

 where, A = Area of square plate = a2

  • If loop is not parallel to the magnetic field 

              Torque \(\vec \tau = IAB \,sin \,\theta\)

             \(\vec \tau = I(\vec A × \vec B)\)      

where,  \(\vec A\)  is the area vector

Note :  Using right hand thumb rule, curl fingers in the direction  of current, then thumb gives the direction of \(\vec A\) which is inside the page.

  • The term \(I \vec A\) is defined as the magnetic dipole moment and denoted by  \(\vec \mu\).

              \( \vec \mu=I \vec A\)

Then , torque \(\vec \tau = \vec \mu × \vec B\)

Unit:

Unit  of magnetic moment \((\vec \mu) = Ampere × m^2\)

  • If a coil of wire contains N loop of same area 

            \(\vec \mu = N I \, \vec A\)

Illustration Questions

Find the torque experienced by a square loop carrying current I= 2 A, having cross section area A = 2 m × 2 m placed  in a magnetic field B = 5 T.                   [\(\alpha = \) 53º ]       [sin 53º = 4/5, sin 37º   = 3/5 ]

A 12 N-m

B 24 N-m

C 18 N-m

D 10 N-m

×

Torque is given as,

    \(\vec \tau = \vec \mu × \vec B\)

     \(\vec \tau = \mu \, B \, sin\, \theta\)                                           [\(\mu = NIA\)]

     \(\vec \tau = N\,I\,A \, B \, sin \,\theta\)

where,

 \(I=\)   Current 

A = Area vector 

B = Magnetic Field 

\(\theta =\) Angle between area vector and magnetic field 

\(\tau = IAB\, sin\theta\)           [N=1]

image

So, torque

\(\tau = 2× 2× 2 × 5 \,sin \,37º\)               

where, 

\(\theta = \) 90º - 53º = 37º 

\(\tau = 40 × \dfrac {3}{5}\)

\(\tau = \)   24 N-m

Find the torque experienced by a square loop carrying current I= 2 A, having cross section area A = 2 m × 2 m placed  in a magnetic field B = 5 T.                   [\(\alpha = \) 53º ]       [sin 53º = 4/5, sin 37º   = 3/5 ]

image
A

12 N-m

.

B

24 N-m

C

18 N-m

D

10 N-m

Option B is Correct

Torque on a Ring moving with Angular Velocity in Uniform Magnetic Field B 

  • Consider a ring of mass m and radius R, having charge q uniformly distributed on its circumference, is moving with angular velocity \(\omega\) in a uniform magnetic field.
  • The ring is placed horizontally and direction of magnetic field is also horizontal.

  • Current flow on ring 

 \(I = \dfrac {q}{T}\)

\(I = \dfrac {q}{\dfrac{2\pi}{\omega}}\)           \( \left[ \therefore T = \dfrac {2\pi}{\omega}\right]\)

 or , \(I = \dfrac {q\omega}{{2\pi}} \)

where,

\(\omega=\) angular velocity

q = charge on ring

  • Magnetic moment on ring 

\(\mu = NIA\)           [N=1]

or, \(\mu = IA\)

or, \(\mu = \dfrac{q\omega}{2\pi} A\)

or,  \(\mu = \dfrac{q\omega}{2\pi} (\pi R^2)\)

or,  \(\mu = \dfrac{\omega R^2 q }{2} \)

  • Torque experienced by ring

\(\vec \tau = \vec \mu × \vec B\)

\(\vec \tau = \dfrac{\omega R^2 qB}{2}\)

 

 

Illustration Questions

Calculate the torque on ring of radius R = 1 m, having charge \( q = 5\,\mu C\) uniformly distributed all over its circumference, rotating with constant angular velocity \(\omega = \) 2 rad/sec in a uniform magnetic field B = 1T, as shown in figure. Given, \(\theta = \) 37º. [sin 37º = 3/5]

A \(8\,\mu\, \text{N-m}\)

B \(6\,\mu\, \text{N-m}\)

C \(3\,\mu\, \text{N-m}\)

D \(2\,\mu\, \text{N-m}\)

×

Current flow on ring 

 \(I = \dfrac {q}{T}\)

\(I = \dfrac {q}{\dfrac{2\pi}{\omega}}\)           \( \left[ \therefore T = \dfrac {2\pi}{\omega}\right]\)

 or , \(I = \dfrac {q\omega}{{2\pi}} \)

where,

\(\omega=\) is angular velocity

q is charge on ring

\(I = \dfrac{5× 10^{-6}× 2}{2\pi}\)

\(I = \dfrac{5}{\pi}\, \mu A\)

Magnetic moment on ring 

\(\mu = NIA\)           [N=1]

or, \(\mu = IA\)

 \(\mu = \dfrac{5}{\pi} × 10^{-6} × \pi (1)^2\)

 \(\mu = 5 \,\mu\, Am^2\)

 

Torque experienced by ring

\(\vec \tau = \vec \mu × \vec B\)

\(\tau = \mu B \,sin\,\theta\)

\(\tau = 5× 10^{-6} × \dfrac{3}{5}\)  

\(\tau = 3 \, \mu\, \text{N-m}\)

Calculate the torque on ring of radius R = 1 m, having charge \( q = 5\,\mu C\) uniformly distributed all over its circumference, rotating with constant angular velocity \(\omega = \) 2 rad/sec in a uniform magnetic field B = 1T, as shown in figure. Given, \(\theta = \) 37º. [sin 37º = 3/5]

image
A

\(8\,\mu\, \text{N-m}\)

.

B

\(6\,\mu\, \text{N-m}\)

C

\(3\,\mu\, \text{N-m}\)

D

\(2\,\mu\, \text{N-m}\)

Option C is Correct

Magnetic Moment Due to 2-Dimensional Square Frame

  • Consider a 2- Dimensional square frame of side 'a', placed in two planes i.e., in x-y plane and y-z plane respectively, carrying current  \(I\) as shown in figure.

  • To calculate magnetic moment due to this frame, we consider frame as two square loops placed in x-y and y-z plane, respectively. 
  • Magnetic moment due to frame in x-y plane

 Magnetic moment     \(\mu _1 = I \vec A\)                                                                     

 where,  \(I\) is current

 \(\vec A \) is area of loop

\(\mu_1= I \,a^2 (\hat k)\)

  • Magnetic moment due to frame in y-z plane

 Magnetic moment  \(\mu _2 = I\vec A \) 

\(\mu _2 = I\,a^2 (\hat i)\)

  • Total magnetic moment 

\(\vec \mu = \vec \mu_1+ \vec \mu_2 \)

\(\vec \mu = I\,a^2 (\hat k ) + I\,a^2 (\hat i )\)

\(\vec \mu = I\,a^2 ( \hat i + \hat k )\)

Illustration Questions

Consider a 2-dimensional square frame of side 'a' = 2 m, placed in x-y and y-z plane respectively, carrying  current \(I = \)2 A, as shown in figure. Calculate the magnetic moment. 

A \(8\,(\hat k + \hat i) \,\text{N-m}^2\)

B \(3 \,(\hat k + \hat i) \,\text{N-m}^2\)

C \(10\,(\hat k + \hat i) \,\text{N-m}^2\)

D \(12 \,(\hat k + \hat i) \,\text{N-m}^2\)

×

Magnetic moment due frame in x-y plane

Magnetic moment     \(\mu _1 = I \vec A\)                                                     

where,

\(I\) is current

\(\vec A \) is area of loop

\(\mu_1= I \,a^2 (\hat k)\)

\(\mu _1 = 2× (2)^2\)

\(\mu _1 = 8 (\hat k)\, Nm^2\)

image

Magnetic moment due to frame in y-z plane

Magnetic moment, \(\mu _2 = I\vec A \)

\(\mu _2 = Ia^2 (\hat i)\)

\(\mu _2 = 2 (2)^2 \hat i\)

\(\mu _2 = 8\,\hat i\, Nm^2\)

image

Total magnetic moment

\(\mu = \mu _1 + \mu_2\)

\(\mu = 8\hat k + 8 \hat i\)

\(\mu = 8(\hat k + \hat i)\, Nm^2\)

Consider a 2-dimensional square frame of side 'a' = 2 m, placed in x-y and y-z plane respectively, carrying  current \(I = \)2 A, as shown in figure. Calculate the magnetic moment. 

image
A

\(8\,(\hat k + \hat i) \,\text{N-m}^2\)

.

B

\(3 \,(\hat k + \hat i) \,\text{N-m}^2\)

C

\(10\,(\hat k + \hat i) \,\text{N-m}^2\)

D

\(12 \,(\hat k + \hat i) \,\text{N-m}^2\)

Option A is Correct

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