Solved Problems on Force between Two Parallel Current Carrying Conductors

Problem 1: Two long parallel wires separated by 0.1 m carry currents of 1A and 2A respectively in opposite directions. A third current-carrying wire parallel to both of them is placed in the same plane such that it feels no net magnetic force. It is placed at a distance of 

Solution: 

We know that the magnetic field due to long straight wire, 

B = μ₀I / 2πx

Therefore, B₁ = B₂ (μ_d / 2πx) =  μ_d2 / 2π(0.1 + x)

Here, i₁ = 1A

         i₂ = 2A 

μ₀×1 / 2πx = μ₀×2 / 2π(0.1 + x) 

2x = (0.1 + x) 

x = 0.1 m

Problem 2: The force per unit length is 10^-3 N on the two current-carrying wires of equal length that are separated by a distance of 2 m and placed parallel to each other. If the current in both the wires is doubled and the distance between the wires is halved, then the force per unit length on the wire will be? 

Solution: 

Force per unit length on both wires f_ab = f_ba = f = 10^-3 N

distance (d) = 2m

The force per unit length on wires is given as,

f_ab = f_ba = f = μ₀I_aI_b / 2πd           —(1)

when the current in both wires is doubled,

I’_a = 2I_a

I’_b = 2I_b

Distance between the wires is halved, 

d’ = d/2

equation (1) can be written as, 

f’_ab = f’_ba = f’ = μ₀I’_aI’_b / 2πd’

f’ = 2 ×  (μ₀×2I_a×2I_b / 2πd)

f’ = 8 × (μ₀×I_a×I_b / 2πd)

f’ = 8f

f’ = 8 × 10-3 N

Problem 3: Two very long wires are placed parallel to each other and separated by a distance of 1m apart. If the current in both the wires is 1A, then the force per unit length on both wires will be: 

Solution: 

Current in both the wires I_a = I_b = 1A

distance (d) = 1m

The force per unit length is given as, 

f_ab = f_ba = f = μ₀I_aI_b / 2πd           —(1)

from equation (1),

f = (μ₀×1×1) / (2π×1)

f = μ₀/2π × 2/2 

f = 2μ₀/4π                —(2)

we know,

μ₀/4π = 10^-7 T – m/A        —(3) 

from equation (2) and (3),

f = 2 × 10^-7 N 

Problem 4: The length of two wires is 0.5 m and the distance between the wires is 1m. If 1 A current is passed in the wires in the same direction, the force per unit length between the wires is: 

Solution: 

Length of two wires (L1 and L2) = 0.5m

Distance between the wires (d) = 1m

Current flowing in each wire (I₁ and  I₂) = 1A 

Magnetic field produced on wire 1 by wire 2 is,

B₂₁ = μ₀I₁ / 2πr = 4π×10^-7×1 / 2π×1 = 2 × 10^-7 T

Magnetic field produced by wire 2 on wire 1 is,

B₁₂ = μ₀I₂ / 2πr = 4π×10-7×1 / 2π×1 = 2 × 10^-7 T

Using Fleming’s left-hand rule,

Force (F₁) and (F₂) is acting on wire 1 and 2 respectively, B₁₂ is directed to the right side and B₂₁ is directed to the left side.  

Using, F = ILB

F₁ = I₁L₁B₁₂ = 1 × 0.5 × (2 x 10^-7) = 1 × 10^-7 N

F₂ = I₂L₂B₂₁ = 1 × 0.5 × (2 x 10^-7) = 1 × 10^-7 N

Since, F1 = F2 and they are directed opposite to one another, the net force is zero.

Hence, the force per unit length between the wires is also zero.  

Problem 5: Wire P carrying current 6 A upward and wire Q is 1m apart from it. If μ₀ = 4π×10^-7 wb A^-1 m^-1 and there is a repulsive force between wire P and Q 1.2×10^-5 N.m^-1. Determine the magnitude and direction of electric current on wire Q. 

Solution: 

Current (I_p) = 6 A

μ₀ = 4π × 10^-7 wb A^-1 m^-1 

Repulsive force (F) = 1.2×10^-5 N m^-1

L = 1 m

Electric current on wire is given by,

F = (μ₀/2π) (I_p I_q/L) 

1.2 × 10^-5 = (4π×10^-7/2π)  (6×I_q/1)

1.2 × 10^-5 = (2×10^-7)  (6×I_q)

1.2 × 10^-5 = (12×10^-7)  (I_q)

1.2  = (12×10^-2)  (I_q)

1.2 = 0.12(I_q)

I_q = 1.2 / 0.12

I_q = 10 A

Problem 6: Wire A and B are 1m apart. Wire P carrying current 1A. If μ₀ = 4π×10^-7 Wb.A^-1.m^-1 and there is an attractive force on each other 10^-7 N.m^-1, determine the magnitude and direction of electric current on wire Q.

Solution: 

Current (I_p) = 1 A

μ₀ = 4π × 10^-7 wbA^-1m^-1

Attractive force (F) = 10^-7 Nm^-1

Length between the wire (L) = 1 m

Electric current on wire is given by,

F = (μ₀/2π) (I_p I_q/L) 

10^-7 = (4π×10^-7/ 2π) (1×I_q / 1)

10^-7 = (2×10^-7) (I_q)

1 = 2(I_q)

I_q = 1/2

I_q = 0.5 A

Moving charges generate an electric field and the rate of flow of charge is known as current. This is the basic concept in Electrostatics. The magnetic effect of electric current is the other important phenomenon related to moving electric charges. Magnetism is generated due to the flow of current. Magnetic fields exert force on the moving charges and at the same time on other magnets, all of which have moving charges. When the charges are stationary, their magnetic field doesn’t affect the magnet but when charges move, they produce magnetic fields that exert force on other magnets.

The movement of charges generates magnetism around a conductor. Generally, magnetism is a property shown by magnets and produced by moving charges, which results in objects being attracted or pushed away.