Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

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
`        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 ?`
10 months ago

Arun
22615 Points
```							Number of turns on the circular coil, n = 100Radius of each turn, r = 1.0 cm = 0.01 mCurrent flowing in the coil, I = 1p AMagnitude of the magnetic field at the centre of the coil is given by the relation,| B | = μ0 2πnI / 4π rWhere,μ0 = 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 THence, the magnitude of the magnetic field is 62.8 × 10–4 T.
```
10 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on General Physics

View all Questions »

### 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