First slide

Magnetic field due to a current carrying circular ring

Question

One metre length of wire carries a constant current. The wire is bent to form a circular loop. The magnetic field at the centre of this loop is B. The same is now bent to form a circular loop of smaller radius to have four turns in the loop. The magnetic field at the centre of this new loop is

Moderate

A

4B
B

16B
C

B/2
D

B/4

Solution

$B^{'} = n^{2} B \Rightarrow B^{'} = {(4)}^{2} B = B^{'} = 16 B$

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