Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
91 views
in Physics by (100k points)
closed by
A short bar magnet placed with its axis at `30^(@)` with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to `4.5 xx 10^(-2)J`. What is the magnitude of magnetic moment of the magnet ?

1 Answer

0 votes
by (101k points)
selected by
 
Best answer
Here `theta = 30^(@), B 0.25 T, r = 4.5 xx 10^(-2) J, M = ?`
As `r = mB = sin theta :. m = (r)/(B sin theta) = (4.5 xx 10^(-2))/(0.25 sin 30^(@))`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...