A circular coil having N turns and radius r carries a current I. It is held in the XZ plane in a magnetic field B\widehat{i}. The torque on the coil due to the magnetic field is :
 

  • Option 1)

    \frac{B\pi r^{2}I}{N}

  • Option 2)

    B \pi r^{2}I\; N

  • Option 3)

    \frac{B r^{2}I}{\pi\: N}

     

  • Option 4)

    Zero

Answers (1)

\vec{B} is along x

M= NIA

M is along y

\tau=M\times B=MB\; \sin \; 90=MB

\tau=NI(\pi\: r^{2})B

\tau=B\pi r^{2}I\; N


 


Option 1)

\frac{B\pi r^{2}I}{N}

Option 2)

B \pi r^{2}I\; N

Option 3)

\frac{B r^{2}I}{\pi\: N}

 

Option 4)

Zero

Preparation Products

Knockout JEE Main April 2021 (One Month)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 14000/- ₹ 6999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Test Series JEE Main April 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions