Informative line

Parallel Current Carrying Long Wires

Practice to calculate the distance of point of zero magnetic field between and outside the two parallel wires.

Force between Two Parallel Current carrying Wires

Case 1

  • Consider two parallel current carrying conductors, as shown in figure.
  • The direction of current in both the conductors is same and both conductors are of length \(\ell\), where \(\ell\) is very large.

  • Force acting on wire 1 due to magnetic field of wire 2 is shown in figure.
  • The magnetic force on wire 1 due to magnetic field of wire 2 

\(\vec{F_{1/2}} = I_1\;\vec{l} × \vec {B_2}\)

where    \(I_1\) is current in wire 1

        \(B_2\) is magnetic field due to current in 2

Note:- \(\vec{l}\) is always taken in the direction of current flow.

  • The magnetic force on wire 2 due to magnetic field of wire 1 

\(\vec{F_{2/1}} = I_2\;\vec{l} × \vec {B_1}\)

Case 2

  • Consider two conductors carrying current in opposite direction, as shown in figure.

  • Force acting on wire 1 due to magnetic field of wire 2

\(\vec{F_{1/2}} = I_1\;\vec{l} × \vec {B_2}\)

  • Force acting on wire 2 due to magnetic field of wire 1

\(\vec{F_{2/1}} = I_2\;\vec{l} × \vec {B_1}\)

 

Conclusion: Parallel conductors carrying current in the same direction attract each other and parallel conductors carrying current in opposite direction repel each other. 

Illustration Questions

Two parallel wires having steady current flow are placed near to each other. The direction of current in wire 1 is as shown in figure. What will be the direction of current in wire 2 if it is given that conductor 1 attracts conductor 2 ? 

A

B

C

D

×

Parallel conductors carrying current in the same direction attract each other and parallel conductors carrying current in opposite direction repel each other. 

image

Since, both conductors attract each other. So in conductor 2, the direction of current will be same as that in conductor 1. 

image

Hence, option (A) is correct.

image

Two parallel wires having steady current flow are placed near to each other. The direction of current in wire 1 is as shown in figure. What will be the direction of current in wire 2 if it is given that conductor 1 attracts conductor 2 ? 

image
A image
B image
C image
D image

Option A is Correct

Force between Two Parallel Wires

  • Consider two straight infinite conductors, 1 and 2, carrying current \(I_1\) and \(I_2\) respectively in the same direction as shown in figure.
  • The magnetic force on length ( \(\ell\)) of wire 1 due to wire 2

\(\vec{F_{1/2}} = I_1\vec{\ell} × \vec{B_2}\)

where

 \(\vec{\ell}\) is vector length of wire 1

 \(\vec{B_2}\) is magnetic field vector of wire 1 due to wire 2

So,  \(F_1 = I_1 {\ell}B_2\)

\(\Rightarrow F_1 = I_1 {\ell}(\dfrac{\mu_0I_2}{2 \pi a}) \;\;\;\;\;[\therefore \;B_2 = \dfrac{\mu_0I_2}{2 \pi a}\,\, for\,\,long\,\,wire]\)

\(\Rightarrow F_1 = \dfrac{\mu_0 I_ 1 I_2}{2 \pi a}{\ell}\)

  • Force per unit length of wire 1 due to wire 2

\(\dfrac{F_1}{\ell} = \dfrac{\mu_0I_1I_2}{2\pi a}\)      

Illustration Questions

Two parallel wires of infinite length each are placed at a distance a = 25 cm apart. The current in wire 1 and wire 2 is \(I_1\) = 2 A and \(I_2\) = 0.5 A, respectively. Calculate force per unit length on wire 1 due to wire 2.  

A 80 N/m

B 0 N/m

C 8 × 10–7 N/m

D 800 N/m

×

Force per unit length on wire 1 due to wire 2 

\(\dfrac{F_1}{\ell} = \dfrac{\mu_0I_1I_2}{2\pi a}\)

image

\(Given:\, I_1=2\,A\,,\,\,\,\,I_2=\,0.5\,A\,,\,\,\,\,a=\,25\,cm\)

\(\dfrac{F_1}{\ell} = \dfrac{\mu_0}{2\pi} × \dfrac{2 × 0.5}{0.25} \)

\(= \dfrac{4 \pi× 10 ^{-7}}{2\pi}× \dfrac{2× .5}{.25}\)

\(\dfrac{F_1}{\ell} = 8 × 10^{-7} \) \(N/m\)

image

Hence, option (C) is correct.

image

Two parallel wires of infinite length each are placed at a distance a = 25 cm apart. The current in wire 1 and wire 2 is \(I_1\) = 2 A and \(I_2\) = 0.5 A, respectively. Calculate force per unit length on wire 1 due to wire 2.  

image
A

80 N/m

.

B

0 N/m

C

8 × 10–7 N/m

D

800 N/m

Option C is Correct

Calculation of Point of Zero Magnetic Field Between the Two Parallel Wires

  • The point of zero magnetic field between two parallel current carrying wires, is possible only when current in both the wires is in same direction.
  • If current in both the wires is in opposite direction, then the magnetic field due to both wires will be in same direction. So, point of zero magnetic field between both the wires is not possible.
  • Case 1 : In the shown figure, the direction of current \(I_1\) and \(I_2\) are in opposite direction, so, the magnetic field due to both wires is inside the page using right hand thumb rule.

  • Case 2 : In the shown figure, the direction of current \(I_1\) and \(I_2\) are in opposite direction, so, the magnetic field due to both wires is outside the page using right hand thumb rule.

  • To calculate point of zero magnetic field, direction of current in both the wires should be same. 
  • Consider two infinitely long wires, placed at a distance 'd' apart, carrying current of \(I_1\) and \(I_2\) in the same direction, as shown in figure.
  • Consider point P as the point of zero magnetic field at a perpendicular distance x from wire 1.
  • Then, the magnetic field at P will be of same magnitude. 

\(|\vec{B_{P_1}}|\)  =  \(|\vec{B_{P_2}}|\)

\(\Rightarrow \dfrac{\mu _0I_1}{2 \pi x}\;\;=\;\;\dfrac{\mu_0I_2}{2 \pi (d-x)}\)

\(\Rightarrow I_1(d-x) = I_2 x\)

\(\Rightarrow I_1d = (I_1+I_2)x\)

\(\Rightarrow x = \dfrac{I_1d}{(I_1+I_2)}\)      

Illustration Questions

Two infinitely long parallel wires carrying current  \(I_1=2\,A\)  and  \(I_2=3\,A\)  in the same direction, are placed at a distance  \(d = 5\, cm\) apart. Calculate the distance of point of zero magnetic field from wire 1 carrying current  \(I_1=2\,A\). 

A 0 cm

B 4 cm

C 2 cm 

D 3 cm 

×

Let point P be the point of zero magnetic field 

then,   \(\dfrac {\mu_0I_1}{2 \pi x}\;\;=\;\;\dfrac{\mu_0 I_2}{2 \pi (d-x)}\)

image

Given:  \(I_1=2\,A\) ,  \(I_2=3\,A\) ,    \(d = 5 \,cm \)

   \(\dfrac{\mu_0I_1}{2 \pi x}\;\;\;=\;\;\;\dfrac{\mu_0I_2}{2 \pi (d - x)}\)

\(\Rightarrow \dfrac {I_1}{x}\;\;=\;\;\dfrac{I_2}{(d-x)}\)

\(\Rightarrow \dfrac {2}{x}\;\;=\;\;\dfrac{3}{(5-x)}\)

\(\Rightarrow\) \(10 - 2x = 3x\)

\(\Rightarrow\) \(5x =10\) 

\(\Rightarrow\) \(x = 2 \) \(cm\)

Hence, option (C) is correct.

Two infinitely long parallel wires carrying current  \(I_1=2\,A\)  and  \(I_2=3\,A\)  in the same direction, are placed at a distance  \(d = 5\, cm\) apart. Calculate the distance of point of zero magnetic field from wire 1 carrying current  \(I_1=2\,A\). 

A

0 cm

.

B

4 cm

C

2 cm 

D

3 cm 

Option C is Correct

Calculation of Point of Zero Magnetic Field Outside the Two Parallel Wires

  • The point of zero magnetic field outside the two infinite long parallel wires can be calculated if current in both the wires is in opposite direction.
  • When the current through both wires is in same direction, point of zero magnetic field is not possible outside the wires as direction of magnetic field due to both the wires is same.
  • Case 1: In the shown figure, the direction of current in both wires \(I_1\) and \(I_2\) are in same direction. The direction of magnetic field to the left of wire 1 is outside the page due to both wires. Also, the direction of magnetic field at the right of wire 2 is inside the page due to both wires. Hence, point of zero magnetic field is not possible outside the wires.

  • Case 2: In the shown figure, the direction of current in both wires \(I_1\) and \(I_2\) are in same direction. The direction of magnetic field to the left of wire 1 is inside the page due to both wires. Also, the direction of magnetic field at the right of wire 2 is outside the page due to both wires. Hence, point of zero magnetic field is not possible outside the wires.

  • To calculate point of zero magnetic field outside the wire, consider two infinitely long wires carrying current  \(I_1\) and \(I_2\)  in the opposite direction, placed at a distance  \(d\) apart, as shown in figure.  

  • Consider a point P as point of zero magnetic field beyond the wire 1 at a perpendicular distance \(x\) as shown in figure.

\(|\vec{B_{p_1}}|\;=\;\ |\vec{B_{p_2}}|\)

\(\Rightarrow \dfrac{\mu_0I_1}{2 \pi x}\;\;=\;\;\dfrac{\mu_0I_2}{2\pi(d+x)}\)

\(\Rightarrow I_1(d+x) = I_2 x\)

\(x = (\dfrac{I_1}{I_2 - I_1}) \;d\)

  • \((I_2 - I_1) \) should be greater than zero as \(x\) is a distance. 

or, \(I_2 - I_1 >0\)

\(\Rightarrow I_2 >I_1\)

  • It means point of zero magnetic field is possible beyond the wire of less current.
  • If current in both the wires, \(I_1 \) and \(I_2\) is same, then 
  • Point of zero magnetic field 

\(x = (\dfrac{I_1}{I_2 - I_1}) \;d\)

  or, \(x = \dfrac {Id}{I-I} = \dfrac{Id}{0} = \infty\)

Thus, point of zero magnetic field is not possible. 

 

Illustration Questions

Two wires carrying current \(I_1 =2\,A\) and \(I_2=3\,A\) are placed at a distance, \(d = 3\,cm\) apart as shown in figure. Calculate the distance of point of zero magnetic field from wire 2. 

A 2 cm at right of wire 2

B 10 cm at right of wire 2

C 6 cm at left of wire 2

D Can't be determined

×

Point of zero magnetic field will be beyond wire of less current .

Hence, point of zero force will be beyond wire 1 since,

\(I_1\) < \(I_2\) 

image image

Hence, at point P magnitude of magnetic field due to wire 1 will be same as magnitude of magnetic field due to wire 2 but in opposite direction 

\(|\vec{B_{p_1}}|\;=\;\ |\vec{B_{p_2}}|\)

\(\Rightarrow \dfrac{\mu_0I_1}{2 \pi x}\;\;=\;\;\dfrac{\mu_0I_2}{2\pi(d+x)}\)

image

Given: \(I_1=2\,A\)  ,   \(I_2=3\,A\) ,   \( d = 3\, cm \)

\(\Rightarrow \dfrac{\mu_0 (2)}{2 \pi x}\;\;=\;\;\dfrac{\mu_0(3)}{2\pi(3 + x)}\)

\(6 + 2x\,= 3x\)

\(x\)\(= 6 \,cm \)

image

Hence, option (C) is correct.

image

Two wires carrying current \(I_1 =2\,A\) and \(I_2=3\,A\) are placed at a distance, \(d = 3\,cm\) apart as shown in figure. Calculate the distance of point of zero magnetic field from wire 2. 

image
A

2 cm at right of wire 2

.

B

10 cm at right of wire 2

C

6 cm at left of wire 2

D

Can't be determined

Option C is Correct

Illustration Questions

Two parallel infinite long wires carrying current  \(I_1=2\,A\)  and  \(I_2=1\,A\) in the same direction, as shown in figure, are placed at a distance \(d=6\,cm\) apart. Calculate the distance and value of current of third wire with direction so that net force on all three wires is zero.   

A 3 cm, 8 A

B 5 cm, 3 A

C 2 cm, 0.67 A

D 2 cm, 3 A

×

Since, parallel conductors are carrying current in same direction. So, both will attract each other.

image

Force per unit length on both the wires, 

\(\dfrac{F}{\ell}\;\;=\;\;\dfrac{\mu_0I_1I_2}{2\pi d}\)

\(\dfrac{F}{\ell} \;\;=\;\;\dfrac{\mu_0(2)(1)}{2\pi(6 × 10 ^{-2})}\)

\(\dfrac{F}{\ell}\;\;=\;\; 6.66 \,\mu N/m\)

image

A third wire carrying current  \(I\) in the opposite direction is placed at a perpendicular distance \(x\) from wire 1.

image image

For  wire 1 to be in equilibrium  

\(\left(\dfrac{F}{\ell}\right)_2\;\;=\;\;\left(\dfrac{F}{ \ell}\right)_3\)

\(\Rightarrow \dfrac{\mu_0I_1I_2}{2 \pi d} \;=\; \dfrac{\mu_0 I_1I}{2\pi(d-x)}\)

\(\Rightarrow I_2(d-x)=Id\)

\(\Rightarrow (I_2-I) d = I_2x\)

\((1-I) 6 = x\)

\(x = 6- 6\;I\)     —(1)

image

For wire 2 to be in equilibrium

\(\left(\dfrac{F}{\ell}\right)_1\;\;=\;\;\left(\dfrac{F}{ \ell}\right)_3\)

\(\Rightarrow \dfrac{\mu_0I_1I_2}{2 \pi d} \;=\; \dfrac{\mu_0 I_2I}{2\pi x}\)

\(\Rightarrow x I_1 = dI\)

\(2 x\;= 6\; I\)

\(I = \dfrac{2x}{6} —(2)\)

image

From (1) and (2)

\(x = 6-6 \left(\dfrac{2}{6}\right)x \;\; \left [ I = \dfrac{2}{6}x \right ] \)

\(x\) = \(6\) –  \(2\)\(x\)

 \(3x\) = \(6\)

\(x = \dfrac{6}{3}\)

\(x\) = \(2\)\(cm\)

image

Current in third wire 

\(I = \dfrac{2}{6} x\)

\(I = \dfrac{2}{6} × 2 = 4/6 \;A\)

\(I = 2/3\;\;= \,0. 67 \;A\)

image

Hence, option (C) is correct.

image

Two parallel infinite long wires carrying current  \(I_1=2\,A\)  and  \(I_2=1\,A\) in the same direction, as shown in figure, are placed at a distance \(d=6\,cm\) apart. Calculate the distance and value of current of third wire with direction so that net force on all three wires is zero.   

image
A

3 cm, 8 A

.

B

5 cm, 3 A

C

2 cm, 0.67 A

D

2 cm, 3 A

Option C is Correct

Illustration Questions

Two infinitely long wires carrying current of same magnitude in same direction, placed \(\ell_1\) distance apart, as shown in figure. A third wire of mass  \(m\)  and length \(\ell\) carrying current \(I\) in opposite direction, is placed among them such that front view of wire seems like equilateral triangle. Find the current in infinitely wire so that third wire levitated in same form of equilateral triangle. 

A \(\dfrac{mg\pi\ell_1}{\mu_0\pi\ell}\)

B \(\dfrac{mgI_1}{I\ell}\)

C \(\dfrac{mg \;2\pi\;\ell_1}{\sqrt3 \;\mu_0\; I \;\ell}\)

D \(\dfrac {mg\;I\;\ell_1}{I_1 \ell}\)

×

As direction of current in third wire is opposite to other two wires, the third wire is repelled by both wire 1 and wire 2. As the current in wire 1 and wire 2 increases, the repulsive force increases and the levitated wire rises to a point at which the weight of wire is once again levitated in equilibrium.   

image image

The horizontal components of the magnetic force on the levitated wire cancel and the vertical component add together

\(\vec{F_{net}} = 2 F_B \;\;cos \;30°\)

\(\vec{F_{net}} = 2\dfrac{\mu_0 \;I_1\;I\;}{2 \pi \ell _1}\ell \;cos \;30°\)

\(\vec{F_{net}} =\dfrac{\sqrt3}{2}\dfrac{\mu_0I_1I}{\pi \ell_1}\ell\)

.

image

For levitation,

\(\vec{F_{net}} = mg\)

\(\dfrac{\sqrt3\mu_0I_1I\ell}{2\pi\ell_1}\;\;=mg\)

\(I_1 = \dfrac{mg\; 2 \pi\;\ell_1}{\sqrt3 \;\mu _0\;I\;\ell}\)

 

image

Hence, option (C) is correct.

image

Two infinitely long wires carrying current of same magnitude in same direction, placed \(\ell_1\) distance apart, as shown in figure. A third wire of mass  \(m\)  and length \(\ell\) carrying current \(I\) in opposite direction, is placed among them such that front view of wire seems like equilateral triangle. Find the current in infinitely wire so that third wire levitated in same form of equilateral triangle. 

image
A

\(\dfrac{mg\pi\ell_1}{\mu_0\pi\ell}\)

.

B

\(\dfrac{mgI_1}{I\ell}\)

C

\(\dfrac{mg \;2\pi\;\ell_1}{\sqrt3 \;\mu_0\; I \;\ell}\)

D

\(\dfrac {mg\;I\;\ell_1}{I_1 \ell}\)

Option C is Correct

Practice Now