Q. 73.8( 40 Votes )

# Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Answer :

Given:

Current in wire A, I_{A} = 8.0 A

Current in wire B, I_{B} = 5.0 A

Distance between the conductors A and B, d = 4 cm

Length of conductor on which we have to calculate force, L = 10cm

Intuitively we can break the problem into two parts,

1) the magnetic field due to wire A, at a distance d.

2) Then we introduce a current carrying conductor B, at distance d and find the force on it due to field created by wire A.

1) Field at distance d due to current I_{A} in conductor A is given by,

…(1)

Where,

B = Magnetic field strength

I = current through the coil

�_{0} is the permeability of free space.

�_{0} = 4 × π × 10^{-7} TmA^{-1}

d = distance of point P

By plugging the values in the equation (1), we get

⇒ |B| = 3.18 × 10^{-6}T

2) Force on a current carrying conductor due to a magnetic field is given by

F = B × I_{B} × L …(2)

Now by putting the values in equation (2) we get,

F = 3.18 × 10^{-6}T × 5A × 0.1 m

⇒ F = 2 × 10^{-5}N

So, the force on the 10 cm section on wire A is 2 × 10^{-5}N. Since the current is flowing in the same direction the force will be attractive in nature.

__Note: The force will be same on both the wires, we can use Newton’s third law of motion to such conclusion.__

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