Two identical coils each of radius `10cm` and having number of turns `100` are lying in perpendicular planes, such that they have common centre. Find the resultant magnetic field at the centre of the coils if they carry currents `3A` and `3sqrt3A` respectively.

Answer

`3*77xx10^-3T`

Explanation

`r_1=r_2=10cm=0*10m`, `n=100`
`I_1=3A`, `I_2=3sqrt3A`

Magnetic field at centre O due to current `I_1` of the coil is
`B_1=(mu_0)/(4pi)(2pinI_1)/(r_1)=(mu_0nI_1)/(2r_1)`
Magnetic field at centre O due to current `I_2` of the other coil is
`B_2=(mu_0)/(4pi)(2pinI_2)/(r_2)=(mu_0nI_2)/(2r_2)`
Here, `vecB_1` and `vecB_2` are perpendicular to each other.
The resultant magnetic field at O is
`B=sqrt(B_1^2+B_2^2)=[((mu_0nI_1)/(2r_1))^2+((mu_0nI_2^2)/(2r_2))^2]^(1//2)`
`=(mu_0n)/(2)[I_1^2/r_1^2+I_2^2/r_2^2]^(1//2)`
`=((4pixx10^-7)xx100)/(2)[(3^2)/((0*1)^2)+((3sqrt3)^2)/((0*1)^2)]^(1//2)`
`=2pixx10^-5xx60=3*77+10^-3T`