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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

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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
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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