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
601 views
in Moving Charges and Magnetism by (26.1k points)
closed by

A Circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

1 Answer

+2 votes
by (26.7k points)
selected by
 
Best answer

Given n = 100, r = 8.0 cm = 8 × 10-2 I = 0.4 A, B = ? 

At the centre of circular coil

B = \(\frac{\mu_0nI}{2r}=\frac{4\pi\times10^{-7}\times100\times0.4}{2\times8\times10^{-2}}\)

= π × 104 T = 3.1 × 10-4T.

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.

...