Asked by akshaygavit07.12dgatl | 26th Sep, 2021, 02:01: PM
Let us assume axial direction of bar magnet is along x-axis and equtorial direction is along y-axis
Axial direction magnetic field at a distance x = ( μo / 4Π ) [ (2M) / x3 ]
Equtorial direction magnetic field at a distance y = ( μo / 4Π ) [ M / y3 ]
If both the magnetic fields are equal, then we have , [ 2 / x3 ] = [ 1 / y3 ]
Hence , we get , ( y / x ) = ( 1 / 21/3 ) = 2 -1/3
Answered by Thiyagarajan K | 26th Sep, 2021, 09:46: PM
- Two small bar magnet are placed in a line at certain distance d apart if the length of each magnet is negligible comapared to d the force between them is inversely proportional to
- Calculate torque and force acting on the current carrying loop.
- pls solve
- Why can't the south pole of the magnetic needle deflect in Ampere swimming Rule??
- Find the magnitude and direction
- A wire of length l is bent in the form of a circular loop with a no of turns and is suspended in a magnetic field of intensity B. Find the expression for the maximum torque produced on the circular loop when a current I is passed through it.
- MUST EVERY MAGNETIC CONFIGURATION HAVE NORTH POLE AND SOUTH POLE AND CAN WE THINK OF MAGNETIC FIELD CONFIGURATION WITH THREE POLES ?
- A MAGNETISED NEEDLE IN A UNIFORM MAGNETIC FIELD EXPERIENCES A TORQUE BUT NO NET FORCE. HOWEVER, AN IRON NAIL NEAR A BAR MAGNET EXPERIENCES A FORCE OF ATTRACTION IN ADDITION TO TORQUE. EXPLAIN
- Cyclotron uses
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number