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
22.4k views
in Physics by (60.8k points)

A circular coil of radius 2.0cm has 500 turns in it and carries a current of 1.0A. Its axis makes an angle of 30° with the uniform magnetic field of magnitude 0.40T that exists in the space. Find the torque acting on the coil.

1 Answer

0 votes
by (150k points)
selected by
 
Best answer

n = 500, r = 0.02m, θ = 30°

i = 1A, B = 4 × 10–1T

i = μ × B = μB Sin30° = niABSin30°

= 500 × 1 × 3.14 × 4 × 10–4 × 4 × 10–1 × (1/2) = 12.56 × 10–2 = 0.1256 ≈ 0.13N-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

...