Ncert Exercises & Solutions

4.7 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.

NEETprep Answer:

Current flowing in wire A, I_A = 8.0 A

Current flowing in wire B, I_B = 5.0 A

Distance between the two wires, r = 4.0 cm = 0.04 m

Length of a section of wire A, L = 10 cm = 0.1 m

Force exerted on length L due to the magnetic field is given as:

F = \frac{μ_{0} I_{A} I_{B} L}{2 πr}

where, μ_{0} = Permeability of free space = 4 π \times 10^{- 7} T m A^{- 1}

F = \frac{4 π \times 10^{- 7} \times 8 \times 5 \times 0.1}{2 π \times 0.04} = 2 \times 10^{- 5} N

The magnitude of force is

2 \times 10^{- 5} N

. This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.

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