Saturday, January 14, 2017

The magnetic induction at center of coil

A Closely wound flat circular coil of 50 turns wound flat circular coil of 50 turns has a diameter of 10 cm carries a current of 4 ampere. Determine the magnetic induction at center of coil.

Solution:




Given:
No. of turns                             :  n              = 50
Diameter of coil                      : d               = 10 cm      = 0.1 m
Current through coil                : I                =  4 A
The magnetic induction at center of current carrying coil can be given as
B = ( μo n I ) / ( 2 r )
Where, r is the radius of coil
B = ( μo n I ) / d
Putting values we get,
B = ( 4 π X 10-7 X 50 X 4 )
B = 2.513 10-4 T
The magnetic induction at a center of coil of 50 turns and 10 cm diameter carrying a current of 4 ampere is

B = 2.513 10-4 T

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