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A circular coil carrying a current I has...

A circular coil carrying a current I has radius R and number of turns N. If all the three, i.e., the current I, radius R and number of turns N are doubled, then, magnetic, field at its centre becomes:

A

Double

B

Half

C

Four times

D

One-fourth

Text Solution

Verified by Experts

The correct Answer is:
A
Magnetic field at the centre coil is B `= ( mu _(0) Ni )/( 2 R)`
if current I, radius R and number of tums N are doubled
then `B .= ( mu _(0) ( 2 N) (2i))/( 2 ( 2R)) = 2 ( mu _(0) Ni )/( 2 R) = 2 B`
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