A rectangular coil of wire of 100 turns and 10×15cm2 size carrying a current of 2 A is in a magnetic field of induction 2×10–3Wb/m2. If the normal to the plane of the coil makes an angle 30° with the field, then the torque on the coil is.

n=100,A=10×15cm2=150×10−4m2B=2×10−3Wb/m2i=2A,θ=30∘τ=iBAnsin⁡θτ=2×2×10−3×150×10−4×100×sin⁡30∘τ=3×10−3Nm

A rectangular coil of wire of 100 turns and 10×15cm² size carrying a current of 2 A is in a magnetic field of induction 2×10^–3Wb/m². If the normal to the plane of the coil makes an angle 30° with the field, then the torque on the coil is.

see full answer

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*

An Intiative by Sri Chaitanya

answer is 1.

(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

$\begin{array}{l} n = 100, A = 10 \times 15 {cm}^{2} \\ = 150 \times 10^{- 4} m^{2} \\ B = 2 \times 10^{- 3} Wb / m^{2} \\ i = 2 A, θ = 30^{\circ} \\ τ = iBAnsin θ \\ τ = 2 \times 2 \times 10^{- 3} \times 150 \times 10^{- 4} \times 100 \times \sin 30^{\circ} \\ τ = 3 \times 10^{- 3} Nm \end{array}$

Watch 3-min video & get full concept clarity

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

Get Expert Academic Guidance – Connect with a Counselor Today!

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