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
0 votes
157 views
in Physics by (90.7k points)
closed by
The final torque on a coil having magnetic moment 25 A m2 in a 5 T uniform external magnetic field, if the coil rotates through an angle of `60^(@)` under the influence of the magnetic field is
A. 216.5Nm
B. 108.25Nm
C. 102.5Nm
D. 258.1Nm

1 Answer

0 votes
by (91.3k points)
selected by
 
Best answer
Correct Answer - B
`|vec(tau)|=|vecmxxvecB|=mBsin theta`
Here, `m=25Am^(2),theta=60^(@),B=5T`
`therefore" "tau=25xx5xxsin 60^(@) or tau=125xx(sqrt3)/(2)=108.25" N m"`

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

...