Q.

A solenoid of 0.4m length with 500 turns carries a current of 3A. A coil of 10 turns and of radius 0.01m carries a current of 0.4A. The torque required to hold the coil with its axis at right angles to that of solenoid in the middle point of it is,

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

6π2×10-7Nm

b

3π2×10-7Nm

c

9π2×10-7Nm

d

12π2×10-7Nm

answer is A.

(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

Field due to solenoid, B=μonI = 4π×10-7×5000.4×3=15π×10-4 T Now, torque experienced by coil in presence of above field Torque =τ=BNiA=BNiπr2=15π×10-4×10×0.4×π×(0.01)2=6π2×10-7Nm
Watch 3-min video & get full concept clarity
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