A straight wire of diameter 1 mm carrying a current of 2 amperes is replaced by another wire of 2 mm diameter. The strength of the magnetic field far away is :
twice the earlier value
one half the earlier value
same as the earlier value
one quarter the earlier value

Question

A straight wire of diameter 1 mm carrying a current of 2 amperes is replaced by another wire of 2 mm diameter. The strength of the magnetic field far away is :
twice the earlier value
one half the earlier value
same as the earlier value
one quarter the earlier value

Toppr Admin · Accepted Answer

Apply Ampere-apos-s Law to the loop outside the wire at the given distance-x3BC-0I-B-2-x3C0-r-x27F9-B-x3BC-0I2-x3C0-rHence the field due to current carrying wire outside is independent of the diameter of the wire-

A straight wire of diameter 1 mm carrying a current of 2 amperes is replaced by another wire of 2 mm diameter. The strength of the magnetic field far away is :twice the earlier valueone half the earlier valuesame as the earlier valueone quarter the earlier value

Apply Ampere's Law to the loop outside the wire at the given distance,μ0I=B(2πr)⟹B=μ0I2πrHence the field due to current carrying wire outside is independent of the diameter of the wire.

A straight wire of diameter 1 mm carrying a current of 2 amperes is replaced by another wire of 2 mm diameter. The strength of the magnetic field far away is :
twice the earlier value
one half the earlier value
same as the earlier value
one quarter the earlier value

Apply Ampere's Law to the loop outside the wire at the given distance,
$μ_{0} I = B (2 π r)$
$⟹ B = \frac{μ_{0} I}{2 π r}$
Hence the field due to current carrying wire outside is independent of the diameter of the wire.