Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A bar magnet of magnetic moment  2.0 \mathrm{~J} / \mathrm{T}  lies aligned with the direction of a uniform magnetic field of  0.25 \mathrm{~T}. what is the amount of work required to turn the magnet so as to align its magnetic moment normal to the field direction?
 

Option: 1

0.125 \mathrm{~J}


Option: 2

0.25 \mathrm{~J}


Option: 3

0.5 \mathrm{~J}


Option: 4

1.0 \mathrm{~J}


Answers (1)

The potential energy of a bar magnet with its magnetic moment \mathbf{M}  inclined at an angle \theta with magnetic field \mathbf{B}  is
\mathrm{ U=-M B \cos \theta }
Potential energy when \mathrm{ \theta=0 }  is
\mathrm{ U_0=-M B \cos 0^{\circ}=-M B }
Potential energy when  \mathrm{ \theta=90^{\circ} } is
\mathrm{ U^{\prime}=-M B \cos 90^{\circ}=0 } 
\mathrm{ \therefore }  Work done  \mathrm{ =U^{\prime}-U_0=0-(-M B)=M B= 2.0 \times 0.25=0.5 \mathrm{~J} }. Hence the correct choice is (c).

Posted by

Ramraj Saini

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question