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.    

Illustration Questions

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 \( B = 2\,T\).

image
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

Illustration Questions

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.

image
A

\(1.2 \,N\)

.

B

\(5\,N\)

C

\(3\,N\)

D

\(0.1\,N\)

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.

Illustration Questions

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 \(\vec{PQ}\) 

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

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

image

Force on wire is given as, 

\(\vec F = I (\vec{PQ} × \vec B)\)

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

\(\vec F = 12\, \hat k\, N\)

image

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. 

image
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

\(I L B =ma\)

\(a = \dfrac{ILB}{m}\)

Illustration Questions

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)

image

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.  

image
A

6 m/sec2

.

B

2 m/sec2

C

8 m/sec2

D

4 m/sec2

Option C 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

Illustration Questions

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.

image
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 ]

Illustration Questions

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.

image

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

image

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 ]

image
A

\(8 \,\pi\, N \)

.

B

\(6 \,\pi\, N \)

C

\(16\,\pi\, N \)

D

\(12\,\pi\, N \)

Option D is Correct

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