Q.

An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current ‘I’ along the same direction as shown in figure. Magnitude of force per unit length on the middle wire ‘B’ is given by

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

a

2μ0I2πd

b

2μ0I2πd

c

μ0I22πd

d

μ0I22πd

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Force between wires A and B = force between wires B and C  ∴FBC=FAB=μoI2l2πd  As F→AB⊥F→BC, net force on wire B,  Fnet =2FBC=2μoI2l2πdFnet =μoI2l2πd  or Fnet l=μoI22πd
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
whats app icon