A current of 1 A flowing on the sides of an equilateral Delta of side `4.5xx10^(-2)` m The magnetic field at the centre of the Delta will be :

Options

A. `4xx10^(-5)Wb//m^(2)`
B. `8xx10^(-5)Wb//m^(2)`
C. `2xx10^(-5)Wb//m^(2)`
D. Zero

Answer

A

Explanation

`B_("net")=3(mu_(0)I)/(4pia)(cos 30^(2)+cos 30^(@))`
`=(3xx(10^(-7))xx1)/((4.5xx10^(-2)))/(2sqrt(3))(2xx(sqrt(3))/2)`
`( :' tan 30^(@)=a/((a//2)), a=1/2 tan 30^(@)=1/(2sqrt(3)))`
`B_("net")=(2xx9xx10^(-5))/4.5=4xx10^(-5)Wb//m^(2)`