Answer :


Given:
Diameter of the copper wire: d = 1.6mm = 1.6× 10-3 m
Current through the wire: I = 20 A
Formula used:
By Ampere’s law for current carrying wire of cross section area,
Where,

B is the magnitude of magnetic field,


μ0 is the permeability of free space and μ0= 4π × 10-7 T mA-1 r is the radius of the wire.


Radius of the wire : r = d/2 = 8× 10-3 m
Substituting we get,

B = 5× 10-3 T
Hence, maximum magnitude of magnetic field due to a current of 20 A is 5× 10-3 T.


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