Informative line

### Force On Current Carrying Wire

Calculate the magnitude of force on a straight wire, carrying current, when placed in a magnetic field and direction of magnetic force on a ring of radius, carrying current, placed in a magnetic field.

# Magnetic Force on a Current Carrying Wire

• Current  carrying wire experiences a force when placed in a magnetic field.
• The force experienced by a wire is the vector sum of individual force exerted on all  charges to form a current.
• Consider a wire carrying current $$I$$ placed in magnetic field as shown in figure.

• By right hand thumb rule, the force on a current carrying wire due to magnetic field will be at left side.

• Magnetic force on a charge q, moving with drift velocity vd  in a magnetic field B is

$$\vec F = q (\vec v_d × \vec B)$$

• The magnetic force on a segment of wire of length L and cross_sectional area is given as

$$\vec F = q (\vec V_d × \vec B) n \,A\,L$$

where, n is the number of charge carrier per unit volume.

• Relation between current and drift velocity is given as

$$I = n\,q \, v_d \,A$$

• Thus magnetic force on wire,

$$\vec F = I (\vec L × \vec B)$$

•  Direction of $$\vec L$$  is taken in the direction of current and $$|\vec L| = L$$ ,i.e., length of segment.

#### Calculate the magnitude of force on a straight wire of length $$L= 2\,m$$, carrying current $$\text I =1\,A$$, when placed in a magnetic field $$B = 2\,T$$.

A 5 N

B 4 N

C 2 N

D 3 N

×

Force $$(\vec F)$$ on a current $$(I)$$ carrying wire of length $$(L)$$ placed in a magnetic field $$(\vec B)$$ is

given as:

$$\vec F = I (\vec L × \vec B )$$

Since, the angle between magnetic field and length is 90º.

$$\vec F =\text I\, \ell\,\text B \,sin\,90º$$

$$\vec F =\text I \,\ell\,\text B$$

Given : $$\text I = 1\,A$$,   $$\text L =2\,m$$, $$B = 2\,T$$

$$\vec F = 1× 2× 2$$

$$\vec F = 4\,N$$

### Calculate the magnitude of force on a straight wire of length $$L= 2\,m$$, carrying current $$\text I =1\,A$$, when placed in a magnetic field $$B = 2\,T$$.

A

5 N

.

B

4 N

C

2 N

D

3 N

Option B is Correct

# Magnetic Force on a Straight Wire Placed at an Angle to the Magnetic Field

• Consider a wire of length $$L$$, placed in a uniform magnetic field $$\vec B$$, making an angle $$\theta$$ with field.

• Magnetic force on a current carrying wire is given as

$$\vec F_B =\text I (\vec L × \vec B )$$

$$|\vec F_B| =\text{ I L B}\, sin\, \theta$$

where,

$$\text L\,sin \,\theta$$  is a  perpendicular  length

$$\text I$$ is current

$$B$$ is magnetic field

#### Find the magnitude of magnetic force caused by magnetic field $$\vec B = 0.2\,T$$ on a wire of length $$L= 1\,m$$, carrying current $$\text I = 1\,A$$ and making an angle $$\theta=30^°$$ with field.

A $$1.2 \,N$$

B $$5\,N$$

C $$3\,N$$

D $$0.1\,N$$

×

Magnetic force on a current carrying wire is given as

$$\vec F_B =\text I (\vec L × \vec B )$$

$$|\vec F_B| =\text{ I L B}\, sin\, \theta$$

where,

$$\text {L sin }\,\theta$$  is a  perpendicular  length

$$\text I$$ is current

$$B$$ is magnetic field

Given : $$\text I = 1\,A$$$$L = 1\,m$$$$\vec B=.2\,T$$$$\theta = 30^°$$

$$\vec F$$ =$$1\times1\times0.2\times\,\text {sin}\,30^\circ$$

$$\vec F=0.1\,N$$

### Find the magnitude of magnetic force caused by magnetic field $$\vec B = 0.2\,T$$ on a wire of length $$L= 1\,m$$, carrying current $$\text I = 1\,A$$ and making an angle $$\theta=30^°$$ with field.

A

$$1.2 \,N$$

.

B

$$5\,N$$

C

$$3\,N$$

D

$$0.1\,N$$

Option D is Correct

# Force on a Sine Shaped Current Carrying Wire

• Consider a sine shaped current ( $$\text I$$) carrying wire, placed in uniform magnetic field (B).

• The length of sine shaped wire

$$\vec{PQ} = 4R$$

• Hence, force on current carrying sine shaped wire

$$\vec F = I (\vec{PQ }× \vec B)$$                      $$\Big[\vec {PQ} = 4R\Big]$$

$$\vec F = I \,4 R\,B\, \hat k \,$$      (outside)

where,

$$\text I$$ is  current

$$R$$ is radius of wire loop

$$B$$ is magnetic field

#### A sine shaped wire of radius $$R= 2\,m$$, carrying current $$\text I =2\,A$$, is placed in magnetic field $$B = 1\,T$$. Find the force due to ring for one full cycle.

A $$8\,\hat k\,N$$

B $$16\,\hat k\,N$$

C $$5\,\hat k\,N$$

D $$4\,\hat k\,N$$

×

The length of sine shaped wire

$$\vec {PQ} = 4R$$

Force on current carrying sine shaped wire

$$\vec F = I (\vec{PQ }× \vec B)$$                      $$\Big[\vec {PQ} = 4R\Big]$$

$$\vec F = I \,4 R\,B\, \hat k \,$$      (outside)

where,

$$\text I$$ is  current

$$R$$ is radius of wire loop

$$B$$ is magnetic field

Given : $$R = 2\,m$$$$\text I =2\,A$$, $$B = 1\,T$$

$$F = 4 × 2× 2 × 1$$

$$F=16\,\hat k\,N$$

### A sine shaped wire of radius $$R= 2\,m$$, carrying current $$\text I =2\,A$$, is placed in magnetic field $$B = 1\,T$$. Find the force due to ring for one full cycle.

A

$$8\,\hat k\,N$$

.

B

$$16\,\hat k\,N$$

C

$$5\,\hat k\,N$$

D

$$4\,\hat k\,N$$

Option B is Correct

# Magnetic Force on a Ring placed near a Strong Magnet

• Consider a strong magnet, placed under a horizontal conducting ring of radius r, carrying current $$\text I$$ as shown in figure.
• A magnetic field B makes an angle $$\theta$$ with the vertical at the ring location.

Here, r is radius of ring and  $$\text I$$ is the current.

• Horizontal components of force will cancel out and only the perpendicular components get added.

• Thus, net force will be in vertical direction.

$$\vec F_{d\ell}=\text I \,d\ell\,\text {B sin}\,\theta$$

$$\vec F = \displaystyle \int \limits _0^{2\pi r} I d\,\ell \, B \, sin\,\theta$$

$$\vec F = I \, B \, sin\,\theta\displaystyle \int \limits _0^{2\pi r} d\,\ell$$

$$\vec F = I B \,sin\, \theta × 2 \pi \,r$$

$$\vec F = 2 \pi \,r \,I B \,sin\, \theta$$       [upward direction ]

#### Calculate the magnitude and direction of magnetic force on a  ring of radius r = 1 m, carrying current  $$\text I=$$  2 A, placed in a magnetic field B = 5 T, created by strong magnet and makes an angle $$\theta =$$ 37º with vertical ring's location.             [sin37º = 3/5 ]

A $$8 \,\pi\, N$$

B $$6 \,\pi\, N$$

C $$16\,\pi\, N$$

D $$12\,\pi\, N$$

×

A magnetic field B makes an angle $$\theta$$ with the vertical at the ring location.

Here, r is radius of ring and  $$\text I$$ is the current.

Horizontal components of force will cancel out and only the perpendicular components get added.

Thus, net force will be in vertical direction.

$$\vec F_{d\ell}=\text I \,d\ell\,\text {B sin}\,\theta$$

$$\vec F = \displaystyle \int \limits _0^{2\pi r} I d\,\ell \, B \, sin\,\theta$$

$$\vec F = I \, B \, sin\,\theta\displaystyle \int \limits _0^{2\pi r} d\,\ell$$

$$\vec F = I B \,sin\, \theta × 2 \pi \,r$$

$$\vec F = 2 \pi \,r \,I B \,sin\, \theta$$       [upward direction ]

Given : $$\text I = 2A$$ ,$$r = 1 \,m$$ , $$B = 5 \,T$$$$\theta =37°$$

$$\vec F = 2 \,\pi × 1 × 2 × 5 \,sin\, 37º$$

$$\vec F = 12 \,\pi\, N$$

### Calculate the magnitude and direction of magnetic force on a  ring of radius r = 1 m, carrying current  $$\text I=$$  2 A, placed in a magnetic field B = 5 T, created by strong magnet and makes an angle $$\theta =$$ 37º with vertical ring's location.             [sin37º = 3/5 ]

A

$$8 \,\pi\, N$$

.

B

$$6 \,\pi\, N$$

C

$$16\,\pi\, N$$

D

$$12\,\pi\, N$$

Option D is Correct

# Force on an Arbitrary Shaped Wire Segment Placed in a Uniform Magnetic Field

• Consider a wire of arbitrary shape, placed in a uniform magnetic field.

• To calculate force on the wire placed in a magnetic field, choose a small element $${d\vec\ell}$$  and analyze force on it.
• The direction of force on this element is outside the page by using right hand thumb rule.
• Force on element $$d\ell$$

$$(\vec F_B)_{d\ell} = I ( {d\vec\ell} × \vec B)$$

• So, the force on whole wire is

$$\vec F_B = I\displaystyle \int_P^Q\, {d\vec\ell} × \vec B$$

$$\vec F_B = I (\vec {PQ}) × \vec B$$

where, P and Q represent the end point of the wire.

#### A wire of arbitrary shape is placed in x - y plane in a uniform magnetic field $$\vec B$$ as shown in figure . Calculate the force on the wire.

A $$6 \, \hat k \, N$$

B $$8 \, \hat k \, N$$

C $$12 \, \hat k \, N$$

D $$10 \, \hat k \, N$$

×

Position Vector $$\vec{PQ}$$

$$\vec {PQ} = (4-0)\hat i + (5-0)\hat j$$

$$\vec {PQ} = 4\hat i + 5\hat j$$

Force on wire is given as,

$$\vec F = I (\vec{PQ} × \vec B)$$

$$\vec F = 1 \left[(4 \hat i + 5 \hat j)× (3 \hat j)\right]$$

$$\vec F = 12\, \hat k\, N$$

### A wire of arbitrary shape is placed in x - y plane in a uniform magnetic field $$\vec B$$ as shown in figure . Calculate the force on the wire.

A

$$6 \, \hat k \, N$$

.

B

$$8 \, \hat k \, N$$

C

$$12 \, \hat k \, N$$

D

$$10 \, \hat k \, N$$

Option C is Correct

# Acceleration of the Current Carrying Rod Placed in Uniform Magnetic Field

• Consider a rod of length $$L$$ and mass m carrying current  $$\text I$$, is placed on a smooth surface in a uniform magnetic field.

• Force on a rod due to magnetic field

$$F = ILB$$   ..........(1)

$$\Big[$$Since, $$\vec B$$ is perpendicular to $$\vec L$$   so,   $$\vec B × \vec L = BL \Big]$$

• By free body diagram of rod

Along y-axis,

N = mg           [Normal force cancels gravitational force]

So, $$\Sigma F = ma$$

F = ma    ..........(2)

On comparing (1) and (2), we get

$$I L B =ma$$

$$a = \dfrac{ILB}{m}$$

#### A uniform rod of length $$L = 2 \,m$$ and mass $$m = 1\,kg$$ carrying current  $$\text I =2\,A$$, is placed in a  uniform magnetic field $$B= 2\,T$$ as shown in figure. Calculate the acceleration of the rod due to magnetic field.

A 6 m/sec2

B 2 m/sec2

C 8 m/sec2

D 4 m/sec2

×

Force on rod due to magnetic field

$$\vec F = I (\vec L × \vec B)$$

[Since, L is perpendicular to B so, $$\vec L × \vec B = LB$$  ]

$$F= ILB$$

where,

$$I$$  is current

$$L$$ is length  of rod

$$B$$ is magnetic field

Given : $$\text I =2\,A$$, $$L = 2\,m$$, $$B = 2\,T$$

$$F = 2 × 2 × 2$$

$$F = 8 \,N$$                      ........(1)

Considering free body diagram of rod

N  = mg

So, F= ma              ......(2)

From (1) and (2),

$$8 = ma$$

$$8 = 1 × a$$

$$a = 8\, m / sec^2$$

### A uniform rod of length $$L = 2 \,m$$ and mass $$m = 1\,kg$$ carrying current  $$\text I =2\,A$$, is placed in a  uniform magnetic field $$B= 2\,T$$ as shown in figure. Calculate the acceleration of the rod due to magnetic field.

A

6 m/sec2

.

B

2 m/sec2

C

8 m/sec2

D

4 m/sec2

Option C is Correct