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# 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. - CBSE (Science) Class 12 - Physics

#### Question

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.

#### Solution

Current flowing in wire A, IA = 8.0 A

Current flowing in wire B, IB = 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:

B=(mu_0 2I_AI_BI)/(4pir)

Where,

mu_0 = Permeability of free space = 4π × 10–7 T m A–1

B=(4pixx10^-7xx2xx8xx5xx0.1)/(4pixx0.04)

=2xx10^-5 N

The magnitude of force is 2 × 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.

Is there an error in this question or solution?

#### APPEARS IN

NCERT Solution for Physics Textbook for Class 12 (2018 to Current)
Chapter 4: Moving Charges and Magnetism
Q: 7 | Page no. 169

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Solution 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. Concept: Magnetic Force.
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