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

$$\vec \mu = N I \, \vec A$$

#### 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 = $$IA$$

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

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

#### 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. 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 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

$$\vec \tau = \vec \mu × \vec B$$

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

#### 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] 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

$$\vec \mu = \vec \mu_1+ \vec \mu_2$$

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

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

#### 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

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

$$\mu _1 = 2× (2)^2$$

$$\mu _1 = 8 (\hat k)\, Nm^2$$ Magnetic moment due to frame in y-z plane

Magnetic moment, $$\mu _2 = I\vec A$$

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

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

$$\mu _2 = 8\,\hat i\, Nm^2$$ 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. 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

# 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.

#### 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] 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

$$\vec \mu = N I \, \vec A$$

#### 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] 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 ] A

12 N-m

.

B

24 N-m

C

18 N-m

D

10 N-m

Option B is Correct