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
        
A verical circular coil of radius 0.1 m and of 10 turns carries a steady current  .When the plane of the coil is normal  to the magnetic meridian, a neutral  point  is observed at the centre of the coil.if    BH=horizontal component of earth's magnetic field =0.314×10^-4 T then find the current  flowing  through the coil.
one month ago

Answers : (2)

Arun
23338 Points
							
Dear student
As we know
B = mu * N* i /2r
hence
I = 2 B* r / 2 N* mu
Now put the value and you will get the amswer 0.5 pi
Hope it helps
one month ago
Vikas TU
9762 Points
							
Dear student 
At neutral point 
Magnetic field become opposite and equal to some other magnetic field , 
Other magnetic field = Horizontal component of earth magnetic field.
Bh = magnetic field at centre of crcular coil = (μ0/4π)*2pie nI /r 
I = (μ0/4π)* rBh /2pie*n 
(10^7*0.1*0.314*10^-4)/ (2*3.14*10) = 0.5A 
Hope this helps  
 
one month ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

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


Course Features

  • 110 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