The magnetic field due to a current carrying loop of radius 3 cm at a point on axis at a distance of 4 cm from its centre of 54 `mu` T. Then , the value of the magnetic field at the centre of the loop is

Options

A. 250`mu` T
B. 150`mu`T
C. 75`mu`T
D. 125`mu`T

Answer

A

Explanation

`B_(axis)=54xx10^(-6)=(muNlr_(2))/(2(r^(2)+x^(2))^(3//2))`
Here, r=3cm =`3xx10^(-2)m`
and x=4 cm `=4xx10^(-2)`m.
From above relation , we get
`muNl=(54xx10^(-2)xx2xx(25xx10^(-4))^(3//2))/((3xx10^(-2))^(2))`
So, `B_("centre")=(muNl)/(2r)=250muT`.