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PHXII04:MOVING CHARGES AND MAGNETISM

362702 $A$ and $B$ are two infinitely long straight parallel conductors. $C$ is another straight conductor of length 1 $m$ kept parallel to A and B as shown in the figure. Then the force experienced by $C$ is
supporting img

1 Towards

A

equal to

0.6 \times 10^{- 5} N

2 Towards

B

equal to

5.4 \times 10^{- 5} N

3 Towards

A

equal to

5.4 \times 10^{- 5} N

4 Towards

B

equal to

0.6 \times 10^{- 5} N

Explanation:

Conductor A attracts conductor $C$ towards $A$ Force towards A
$\begin{aligned} = \frac{μ_{0}}{2 π} \frac{I_{1} I_{2} l}{r} = \frac{μ_{0}}{2 π} \frac{2 \times 3 \times 1}{0.05} = \frac{3 μ_{0}}{π} \times 20 \\ = \frac{60 \times 4 π \times 10^{- 7}}{π} = 240 \times 10^{- 7} N \end{aligned}$
Again force towards
$\begin{aligned} B = \frac{μ_{0}}{2 π} \times \frac{3 \times 4 \times 1}{0.08} = \frac{3 μ_{0}}{π} \times 25 \\ = \frac{3 \times 4 π \times 10^{- 7}}{π} \times 25 = 300 \times 10^{- 7} N \end{aligned}$
$∴$ Resultant force towards B $= (300 - 240)$
$10^{- 7} N$
$= 0.6 \times 10^{- 5} N$ .

KCET - 2007

PHXII04:MOVING CHARGES AND MAGNETISM

362703 Three long, straight parallel wires carrying current, are arranged as shown in figure. The force experienced by a 25 $c m$ length of wire $C$ is
supporting img

1

10^{3} N

2

2.5 \times 10^{3} N

3 zero

4

1.5 \times 10^{- 3} N

Explanation:

Force on wire $C$ due to wire $D$
$F_{A} = 10^{- 7} \times \frac{2 \times 30 \times 10}{3 \times 10^{- 2}} \times 25 \times 10^{- 2} = 5 \times 10^{- 4} N$
(towards right)
supporting img
Force on wire $C$ due to wire $B$
$F_{B} = 10^{- 7} \times \frac{2 \times 20 \times 10}{2 \times 10^{- 2}} \times 25 \times 10^{- 2} = 5 \times 10^{- 4} N$
(towards left)
$\Rightarrow$ Net force on wire $C$ is $F_{n e t} = F_{A} - F_{B} = 0$

PHXII04:MOVING CHARGES AND MAGNETISM

362704 Two parallel wires 1 $m$ apart carry current of 1 $A$ and 3 $A$ respectively in opposite directions. The force per unit length acting between these two wires is

1

6 \times 10^{- 7} N m^{- 1}

attractive

2

6 \times 10^{- 5} N m^{- 1}

attractive

3

6 \times 10^{- 7} N m^{- 1}

repulsive

4

6 \times 10^{- 5} N m^{- 1}

repulsive

Explanation:

supporting img
The force per unit length acting between two parallel wires carrying currents $I_{1}$ and $I_{2}$ and placed a distance $d$ apart is
$f = \frac{μ_{0} I_{1} I_{2}}{2 π d}$
Here, $I_{1} = 1 A, I_{2} = 3, A, d = 1 m$
$μ_{0} = 4 π \times 10^{- 7} T m A^{- 1}$
$∴ f = \frac{(4 π \times 10^{- 7} T m A^{- 1}) (1 A) (3 A)}{2 π (1 m)} = 6 \times 10^{- 7} N m^{- 1}$
As the currents are in opposite directions, so $f$ is repulsive

KCET - 2015

PHXII04:MOVING CHARGES AND MAGNETISM

362705 Assertion :
Two electron beams moving parallel come close each other
Reason :
Wires carrying current in opposite direction attracy each other

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII04:MOVING CHARGES AND MAGNETISM

362706 An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current $I$ along the same direction as shown in fig. Magnitude of force per unit length on the middle wire $B^{'}$ is given by :-
supporting img

1

\frac{2 μ_{0} i^{2}}{π d}

2

\frac{\sqrt{2} μ_{0} i^{2}}{π d}

3

\frac{μ_{0} {\dot{i}}^{2}}{\sqrt{2} π d}

4

\frac{μ_{0} i^{2}}{2 π d}

Explanation:

$F = \frac{μ_{0} i_{1} i_{2}}{2 π d} =$ force per unit length
$F_{1} = \frac{(μ_{0} i) i}{2 π d} = \frac{μ_{0} i^{2}}{2 π d} = F_{2}$
supporting img
$F_{net} = \sqrt{F_{1}^{2} + F_{2}^{2}} = \frac{μ_{0} {\dot{i}}^{2}}{\sqrt{2} π d}$

NEET - 2017

PHXII04:MOVING CHARGES AND MAGNETISM

362702 $A$ and $B$ are two infinitely long straight parallel conductors. $C$ is another straight conductor of length 1 $m$ kept parallel to A and B as shown in the figure. Then the force experienced by $C$ is
supporting img

1 Towards

A

equal to

0.6 \times 10^{- 5} N

2 Towards

B

equal to

5.4 \times 10^{- 5} N

3 Towards

A

equal to

5.4 \times 10^{- 5} N

4 Towards

B

equal to

0.6 \times 10^{- 5} N

Explanation:

Conductor A attracts conductor $C$ towards $A$ Force towards A
$\begin{aligned} = \frac{μ_{0}}{2 π} \frac{I_{1} I_{2} l}{r} = \frac{μ_{0}}{2 π} \frac{2 \times 3 \times 1}{0.05} = \frac{3 μ_{0}}{π} \times 20 \\ = \frac{60 \times 4 π \times 10^{- 7}}{π} = 240 \times 10^{- 7} N \end{aligned}$
Again force towards
$\begin{aligned} B = \frac{μ_{0}}{2 π} \times \frac{3 \times 4 \times 1}{0.08} = \frac{3 μ_{0}}{π} \times 25 \\ = \frac{3 \times 4 π \times 10^{- 7}}{π} \times 25 = 300 \times 10^{- 7} N \end{aligned}$
$∴$ Resultant force towards B $= (300 - 240)$
$10^{- 7} N$
$= 0.6 \times 10^{- 5} N$ .

KCET - 2007

PHXII04:MOVING CHARGES AND MAGNETISM

362703 Three long, straight parallel wires carrying current, are arranged as shown in figure. The force experienced by a 25 $c m$ length of wire $C$ is
supporting img

1

10^{3} N

2

2.5 \times 10^{3} N

3 zero

4

1.5 \times 10^{- 3} N

Explanation:

Force on wire $C$ due to wire $D$
$F_{A} = 10^{- 7} \times \frac{2 \times 30 \times 10}{3 \times 10^{- 2}} \times 25 \times 10^{- 2} = 5 \times 10^{- 4} N$
(towards right)
supporting img
Force on wire $C$ due to wire $B$
$F_{B} = 10^{- 7} \times \frac{2 \times 20 \times 10}{2 \times 10^{- 2}} \times 25 \times 10^{- 2} = 5 \times 10^{- 4} N$
(towards left)
$\Rightarrow$ Net force on wire $C$ is $F_{n e t} = F_{A} - F_{B} = 0$

PHXII04:MOVING CHARGES AND MAGNETISM

362704 Two parallel wires 1 $m$ apart carry current of 1 $A$ and 3 $A$ respectively in opposite directions. The force per unit length acting between these two wires is

1

6 \times 10^{- 7} N m^{- 1}

attractive

2

6 \times 10^{- 5} N m^{- 1}

attractive

3

6 \times 10^{- 7} N m^{- 1}

repulsive

4

6 \times 10^{- 5} N m^{- 1}

repulsive

Explanation:

supporting img
The force per unit length acting between two parallel wires carrying currents $I_{1}$ and $I_{2}$ and placed a distance $d$ apart is
$f = \frac{μ_{0} I_{1} I_{2}}{2 π d}$
Here, $I_{1} = 1 A, I_{2} = 3, A, d = 1 m$
$μ_{0} = 4 π \times 10^{- 7} T m A^{- 1}$
$∴ f = \frac{(4 π \times 10^{- 7} T m A^{- 1}) (1 A) (3 A)}{2 π (1 m)} = 6 \times 10^{- 7} N m^{- 1}$
As the currents are in opposite directions, so $f$ is repulsive

KCET - 2015

PHXII04:MOVING CHARGES AND MAGNETISM

362705 Assertion :
Two electron beams moving parallel come close each other
Reason :
Wires carrying current in opposite direction attracy each other

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII04:MOVING CHARGES AND MAGNETISM

362706 An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current $I$ along the same direction as shown in fig. Magnitude of force per unit length on the middle wire $B^{'}$ is given by :-
supporting img

1

\frac{2 μ_{0} i^{2}}{π d}

2

\frac{\sqrt{2} μ_{0} i^{2}}{π d}

3

\frac{μ_{0} {\dot{i}}^{2}}{\sqrt{2} π d}

4

\frac{μ_{0} i^{2}}{2 π d}

Explanation:

$F = \frac{μ_{0} i_{1} i_{2}}{2 π d} =$ force per unit length
$F_{1} = \frac{(μ_{0} i) i}{2 π d} = \frac{μ_{0} i^{2}}{2 π d} = F_{2}$
supporting img
$F_{net} = \sqrt{F_{1}^{2} + F_{2}^{2}} = \frac{μ_{0} {\dot{i}}^{2}}{\sqrt{2} π d}$

NEET - 2017

PHXII04:MOVING CHARGES AND MAGNETISM

362702 $A$ and $B$ are two infinitely long straight parallel conductors. $C$ is another straight conductor of length 1 $m$ kept parallel to A and B as shown in the figure. Then the force experienced by $C$ is
supporting img

1 Towards

A

equal to

0.6 \times 10^{- 5} N

2 Towards

B

equal to

5.4 \times 10^{- 5} N

3 Towards

A

equal to

5.4 \times 10^{- 5} N

4 Towards

B

equal to

0.6 \times 10^{- 5} N

Explanation:

Conductor A attracts conductor $C$ towards $A$ Force towards A
$\begin{aligned} = \frac{μ_{0}}{2 π} \frac{I_{1} I_{2} l}{r} = \frac{μ_{0}}{2 π} \frac{2 \times 3 \times 1}{0.05} = \frac{3 μ_{0}}{π} \times 20 \\ = \frac{60 \times 4 π \times 10^{- 7}}{π} = 240 \times 10^{- 7} N \end{aligned}$
Again force towards
$\begin{aligned} B = \frac{μ_{0}}{2 π} \times \frac{3 \times 4 \times 1}{0.08} = \frac{3 μ_{0}}{π} \times 25 \\ = \frac{3 \times 4 π \times 10^{- 7}}{π} \times 25 = 300 \times 10^{- 7} N \end{aligned}$
$∴$ Resultant force towards B $= (300 - 240)$
$10^{- 7} N$
$= 0.6 \times 10^{- 5} N$ .

KCET - 2007

PHXII04:MOVING CHARGES AND MAGNETISM

362703 Three long, straight parallel wires carrying current, are arranged as shown in figure. The force experienced by a 25 $c m$ length of wire $C$ is
supporting img

1

10^{3} N

2

2.5 \times 10^{3} N

3 zero

4

1.5 \times 10^{- 3} N

Explanation:

Force on wire $C$ due to wire $D$
$F_{A} = 10^{- 7} \times \frac{2 \times 30 \times 10}{3 \times 10^{- 2}} \times 25 \times 10^{- 2} = 5 \times 10^{- 4} N$
(towards right)
supporting img
Force on wire $C$ due to wire $B$
$F_{B} = 10^{- 7} \times \frac{2 \times 20 \times 10}{2 \times 10^{- 2}} \times 25 \times 10^{- 2} = 5 \times 10^{- 4} N$
(towards left)
$\Rightarrow$ Net force on wire $C$ is $F_{n e t} = F_{A} - F_{B} = 0$

PHXII04:MOVING CHARGES AND MAGNETISM

362704 Two parallel wires 1 $m$ apart carry current of 1 $A$ and 3 $A$ respectively in opposite directions. The force per unit length acting between these two wires is

1

6 \times 10^{- 7} N m^{- 1}

attractive

2

6 \times 10^{- 5} N m^{- 1}

attractive

3

6 \times 10^{- 7} N m^{- 1}

repulsive

4

6 \times 10^{- 5} N m^{- 1}

repulsive

Explanation:

supporting img
The force per unit length acting between two parallel wires carrying currents $I_{1}$ and $I_{2}$ and placed a distance $d$ apart is
$f = \frac{μ_{0} I_{1} I_{2}}{2 π d}$
Here, $I_{1} = 1 A, I_{2} = 3, A, d = 1 m$
$μ_{0} = 4 π \times 10^{- 7} T m A^{- 1}$
$∴ f = \frac{(4 π \times 10^{- 7} T m A^{- 1}) (1 A) (3 A)}{2 π (1 m)} = 6 \times 10^{- 7} N m^{- 1}$
As the currents are in opposite directions, so $f$ is repulsive

KCET - 2015

PHXII04:MOVING CHARGES AND MAGNETISM

362705 Assertion :
Two electron beams moving parallel come close each other
Reason :
Wires carrying current in opposite direction attracy each other

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII04:MOVING CHARGES AND MAGNETISM

362706 An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current $I$ along the same direction as shown in fig. Magnitude of force per unit length on the middle wire $B^{'}$ is given by :-
supporting img

1

\frac{2 μ_{0} i^{2}}{π d}

2

\frac{\sqrt{2} μ_{0} i^{2}}{π d}

3

\frac{μ_{0} {\dot{i}}^{2}}{\sqrt{2} π d}

4

\frac{μ_{0} i^{2}}{2 π d}

Explanation:

$F = \frac{μ_{0} i_{1} i_{2}}{2 π d} =$ force per unit length
$F_{1} = \frac{(μ_{0} i) i}{2 π d} = \frac{μ_{0} i^{2}}{2 π d} = F_{2}$
supporting img
$F_{net} = \sqrt{F_{1}^{2} + F_{2}^{2}} = \frac{μ_{0} {\dot{i}}^{2}}{\sqrt{2} π d}$

NEET - 2017

PHXII04:MOVING CHARGES AND MAGNETISM

362702 $A$ and $B$ are two infinitely long straight parallel conductors. $C$ is another straight conductor of length 1 $m$ kept parallel to A and B as shown in the figure. Then the force experienced by $C$ is
supporting img

1 Towards

A

equal to

0.6 \times 10^{- 5} N

2 Towards

B

equal to

5.4 \times 10^{- 5} N

3 Towards

A

equal to

5.4 \times 10^{- 5} N

4 Towards

B

equal to

0.6 \times 10^{- 5} N

Explanation:

Conductor A attracts conductor $C$ towards $A$ Force towards A
$\begin{aligned} = \frac{μ_{0}}{2 π} \frac{I_{1} I_{2} l}{r} = \frac{μ_{0}}{2 π} \frac{2 \times 3 \times 1}{0.05} = \frac{3 μ_{0}}{π} \times 20 \\ = \frac{60 \times 4 π \times 10^{- 7}}{π} = 240 \times 10^{- 7} N \end{aligned}$
Again force towards
$\begin{aligned} B = \frac{μ_{0}}{2 π} \times \frac{3 \times 4 \times 1}{0.08} = \frac{3 μ_{0}}{π} \times 25 \\ = \frac{3 \times 4 π \times 10^{- 7}}{π} \times 25 = 300 \times 10^{- 7} N \end{aligned}$
$∴$ Resultant force towards B $= (300 - 240)$
$10^{- 7} N$
$= 0.6 \times 10^{- 5} N$ .

KCET - 2007

PHXII04:MOVING CHARGES AND MAGNETISM

362703 Three long, straight parallel wires carrying current, are arranged as shown in figure. The force experienced by a 25 $c m$ length of wire $C$ is
supporting img

1

10^{3} N

2

2.5 \times 10^{3} N

3 zero

4

1.5 \times 10^{- 3} N

Explanation:

Force on wire $C$ due to wire $D$
$F_{A} = 10^{- 7} \times \frac{2 \times 30 \times 10}{3 \times 10^{- 2}} \times 25 \times 10^{- 2} = 5 \times 10^{- 4} N$
(towards right)
supporting img
Force on wire $C$ due to wire $B$
$F_{B} = 10^{- 7} \times \frac{2 \times 20 \times 10}{2 \times 10^{- 2}} \times 25 \times 10^{- 2} = 5 \times 10^{- 4} N$
(towards left)
$\Rightarrow$ Net force on wire $C$ is $F_{n e t} = F_{A} - F_{B} = 0$

PHXII04:MOVING CHARGES AND MAGNETISM

362704 Two parallel wires 1 $m$ apart carry current of 1 $A$ and 3 $A$ respectively in opposite directions. The force per unit length acting between these two wires is

1

6 \times 10^{- 7} N m^{- 1}

attractive

2

6 \times 10^{- 5} N m^{- 1}

attractive

3

6 \times 10^{- 7} N m^{- 1}

repulsive

4

6 \times 10^{- 5} N m^{- 1}

repulsive

Explanation:

supporting img
The force per unit length acting between two parallel wires carrying currents $I_{1}$ and $I_{2}$ and placed a distance $d$ apart is
$f = \frac{μ_{0} I_{1} I_{2}}{2 π d}$
Here, $I_{1} = 1 A, I_{2} = 3, A, d = 1 m$
$μ_{0} = 4 π \times 10^{- 7} T m A^{- 1}$
$∴ f = \frac{(4 π \times 10^{- 7} T m A^{- 1}) (1 A) (3 A)}{2 π (1 m)} = 6 \times 10^{- 7} N m^{- 1}$
As the currents are in opposite directions, so $f$ is repulsive

KCET - 2015

PHXII04:MOVING CHARGES AND MAGNETISM

362705 Assertion :
Two electron beams moving parallel come close each other
Reason :
Wires carrying current in opposite direction attracy each other

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII04:MOVING CHARGES AND MAGNETISM

362706 An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current $I$ along the same direction as shown in fig. Magnitude of force per unit length on the middle wire $B^{'}$ is given by :-
supporting img

1

\frac{2 μ_{0} i^{2}}{π d}

2

\frac{\sqrt{2} μ_{0} i^{2}}{π d}

3

\frac{μ_{0} {\dot{i}}^{2}}{\sqrt{2} π d}

4

\frac{μ_{0} i^{2}}{2 π d}

Explanation:

$F = \frac{μ_{0} i_{1} i_{2}}{2 π d} =$ force per unit length
$F_{1} = \frac{(μ_{0} i) i}{2 π d} = \frac{μ_{0} i^{2}}{2 π d} = F_{2}$
supporting img
$F_{net} = \sqrt{F_{1}^{2} + F_{2}^{2}} = \frac{μ_{0} {\dot{i}}^{2}}{\sqrt{2} π d}$

NEET - 2017

PHXII04:MOVING CHARGES AND MAGNETISM

362702 $A$ and $B$ are two infinitely long straight parallel conductors. $C$ is another straight conductor of length 1 $m$ kept parallel to A and B as shown in the figure. Then the force experienced by $C$ is
supporting img

1 Towards

A

equal to

0.6 \times 10^{- 5} N

2 Towards

B

equal to

5.4 \times 10^{- 5} N

3 Towards

A

equal to

5.4 \times 10^{- 5} N

4 Towards

B

equal to

0.6 \times 10^{- 5} N

Explanation:

Conductor A attracts conductor $C$ towards $A$ Force towards A
$\begin{aligned} = \frac{μ_{0}}{2 π} \frac{I_{1} I_{2} l}{r} = \frac{μ_{0}}{2 π} \frac{2 \times 3 \times 1}{0.05} = \frac{3 μ_{0}}{π} \times 20 \\ = \frac{60 \times 4 π \times 10^{- 7}}{π} = 240 \times 10^{- 7} N \end{aligned}$
Again force towards
$\begin{aligned} B = \frac{μ_{0}}{2 π} \times \frac{3 \times 4 \times 1}{0.08} = \frac{3 μ_{0}}{π} \times 25 \\ = \frac{3 \times 4 π \times 10^{- 7}}{π} \times 25 = 300 \times 10^{- 7} N \end{aligned}$
$∴$ Resultant force towards B $= (300 - 240)$
$10^{- 7} N$
$= 0.6 \times 10^{- 5} N$ .

KCET - 2007

PHXII04:MOVING CHARGES AND MAGNETISM

362703 Three long, straight parallel wires carrying current, are arranged as shown in figure. The force experienced by a 25 $c m$ length of wire $C$ is
supporting img

1

10^{3} N

2

2.5 \times 10^{3} N

3 zero

4

1.5 \times 10^{- 3} N

Explanation:

Force on wire $C$ due to wire $D$
$F_{A} = 10^{- 7} \times \frac{2 \times 30 \times 10}{3 \times 10^{- 2}} \times 25 \times 10^{- 2} = 5 \times 10^{- 4} N$
(towards right)
supporting img
Force on wire $C$ due to wire $B$
$F_{B} = 10^{- 7} \times \frac{2 \times 20 \times 10}{2 \times 10^{- 2}} \times 25 \times 10^{- 2} = 5 \times 10^{- 4} N$
(towards left)
$\Rightarrow$ Net force on wire $C$ is $F_{n e t} = F_{A} - F_{B} = 0$

PHXII04:MOVING CHARGES AND MAGNETISM

362704 Two parallel wires 1 $m$ apart carry current of 1 $A$ and 3 $A$ respectively in opposite directions. The force per unit length acting between these two wires is

1

6 \times 10^{- 7} N m^{- 1}

attractive

2

6 \times 10^{- 5} N m^{- 1}

attractive

3

6 \times 10^{- 7} N m^{- 1}

repulsive

4

6 \times 10^{- 5} N m^{- 1}

repulsive

Explanation:

supporting img
The force per unit length acting between two parallel wires carrying currents $I_{1}$ and $I_{2}$ and placed a distance $d$ apart is
$f = \frac{μ_{0} I_{1} I_{2}}{2 π d}$
Here, $I_{1} = 1 A, I_{2} = 3, A, d = 1 m$
$μ_{0} = 4 π \times 10^{- 7} T m A^{- 1}$
$∴ f = \frac{(4 π \times 10^{- 7} T m A^{- 1}) (1 A) (3 A)}{2 π (1 m)} = 6 \times 10^{- 7} N m^{- 1}$
As the currents are in opposite directions, so $f$ is repulsive

KCET - 2015

PHXII04:MOVING CHARGES AND MAGNETISM

362705 Assertion :
Two electron beams moving parallel come close each other
Reason :
Wires carrying current in opposite direction attracy each other

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII04:MOVING CHARGES AND MAGNETISM

362706 An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current $I$ along the same direction as shown in fig. Magnitude of force per unit length on the middle wire $B^{'}$ is given by :-
supporting img

1

\frac{2 μ_{0} i^{2}}{π d}

2

\frac{\sqrt{2} μ_{0} i^{2}}{π d}

3

\frac{μ_{0} {\dot{i}}^{2}}{\sqrt{2} π d}

4

\frac{μ_{0} i^{2}}{2 π d}

Explanation:

$F = \frac{μ_{0} i_{1} i_{2}}{2 π d} =$ force per unit length
$F_{1} = \frac{(μ_{0} i) i}{2 π d} = \frac{μ_{0} i^{2}}{2 π d} = F_{2}$
supporting img
$F_{net} = \sqrt{F_{1}^{2} + F_{2}^{2}} = \frac{μ_{0} {\dot{i}}^{2}}{\sqrt{2} π d}$

NEET - 2017