What is the magnetic field along the axis and equatorial line of a bar magnet ?

Explanation

`B= (mu_0I)/(2a) .` Given `2pi a = 88 xx 10^(-2) " " therefore a= (88xx 10^(-2))/(2pi)` , I = 1A
`therefore B = (4pi xx 10^(-7) xx 1)/(2 xx 88/(2pi) xx 10^(-7)) = (pi^2 xx 10^(-7))/(22 xx 10^(-2))`
`therefore B= (22)/(7) xx (22)/(7) xx (10^(-5))/(22) ~~ 0.45 xx 10^(-5) T`
b. Zero
c. `~~ 1.8 xx 10^(-5) T`