MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 12
        

Consider a tightly wound 100 turns of a coil of radius 1 cm carrying a current of 1A . What is the magnitude of the magnetic field at the center of the coil ?

6 months ago

Answers : (1)

Arun
21015 Points
							

Number of turns on the circular coil, n = 100

Radius of each turn, r = 1.0 cm = 0.01 m

Current flowing in the coil, I = 1p A

Magnitude of the magnetic field at the centre of the coil is given by the relation,

| B | = μ0 2πnI / 4π r

Where,

μ= Permeability of free space

= 4π × 10–7 T m A–1

| B | = 4π x 10-7 x 2π x 100 x 1 / 4π x 0.01

       = 62.8 x 10-4 T

Hence, the magnitude of the magnetic field is 62.8 × 10–4 T.

6 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 18 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details