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The magnetic field at the centre of curr...

The magnetic field at the centre of current carrying coil is `B_(0)`. If its radius is reduced to half keeping current the same, then magnetic field at its centre become

A

`B_(0)`

B

`2B_(0)`

C

`4B_(0)`

D

`((B_(0))/(2))`

Text Solution

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The correct Answer is:
B
We know that, The magnetic field at the centre of current carrying coll.
`B_(0)= (mu_(0)l)/(2r)`
Now according to question,
`B= ((mu_(0)i)/(2((r)/(2)))) = 2 ((mu_(0)i)/(2r))= 2 B_(0)`
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