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 rectangular coil pqrs of N turns length L and breadth B and carrying a current i is placed in a uniform magnetic field B with its longest side perpendicular and shorter sides parallel to the field as shown in the figure .compute the torque on the coil .the normal to its plane makes an angle Alpha with the field
26 days ago

Answers : (1)

Arun
16176 Points
							
The area of cross-section of the coil is A = lb
 
Now, the magnetic moment of the coil is
 
m = IA = Ilb
 
Now, we know that the torque acting on the coil is given as
 
T = m x B = mb sin theta
 
Here, θ is the angle between the magnetic field and the normal to the plane of the coil.
 
Now, the magnetic field is directed perpendicular to the coil. Also, the normal to the coil will be perpendicular to the coil.
 
So, the angle between the normal and the field is 0.
 
Hence, the torque acting on the coil is zero.
26 days ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 276 off

COUPON CODE: SELF10


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


Get Extra Rs. 297 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

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