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The magnetic field due to current carryi...

The magnetic field due to current carrying circular coil having n turns is

A

`n^(2)` times as large as that produced by a single turn

B

n times as large as that produced by a single turn

C

`(1)/(2)` -times as large as that produced by a single turn

D

will be same as that produced by a single turn

Text Solution

Verified by Experts

The correct Answer is:
B
Magnetic field due to a single turn of wire `B = ( mu_(0) l)/(L)` (Where , L is the length of wire )
Now, when this wire have n turns, then its magnetic field will be.
` B = (mu_(0) nl)/(L) = mu_(0) NI`
(Here, Nis number of turns per unit length)
` rArr B. = nB=n ` times of field of single turn.
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