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
rohit rathi Grade: 12
        

A long thin walled pipe of radius R carries current I along its length.The current density is uniform over circumference of pipe.The magnetic field at the centre of pipe due to quarter portion of pipe?-Please answer and explain.....

6 years ago

Answers : (2)

vikas askiitian expert
510 Points
										

total current through circumfrence  = I


 current through unit segment of circumfrenece = I/2piR                   (current dencity)


length of quater of segment = 2piR/4 = piR/2


 current through quater segment (piR/2)= (I/2piR)(piR/2)


                                                        =I/4


now we have , B.dl = uoI(current threading)


                    B(2PiR) = uo(I/4)


                    B = uoI/8piR

6 years ago
Jerin
19 Points
										I current - 2πdI - dΦdI=IdΦ/2πdB=u0dI/2πR     =u0IdΦ/4ππRB=integration of dB    =u0I[i+j]/4ππRR   =u0I√2/4ππRR
										
3 days ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: Rs. 2,756
  • View Details
  • Complete Physics Course - Class 11
  • OFFERED PRICE: Rs. 2,968
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

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