Two circular coils of mean radii 0.1 m and 0.5 m consisting of 5 turns and 10 turns respectively are arranged concentric to one another with their planes at right angles to each other. If a current of 2A is passed through each of them, calculate the magnitude of the resultant magnetic field at their common centre.

Explanation

Given : `r_(1)=0.10 m, r_(2)=0.05m, n_(1)=5, n_(2)=10, I=2A`
We know that, B at the centre of the circular conductor carrying current
`" "B=mu_(0)/(4pi)(2pinI)/r""tesla`
Hence, `" "B_(1)=(10^(-7) times 2 times 3.142 times 5 times 2)/(0.10)`
`" "B_(1)=6.284 times 10^(-5)T`
Similarly, `" "B_(2)=(10^(-7) times 2 times 3.142 times 10 times 2)/(0.05)`
`" "B_(2)=25.136 times 10^(-5)T`
Resultant field, `" "B=sqrt(B_(1)^(2)+B_(2)^(2))`
`" "=10^(-5)sqrt((284)^(2)+(25.136)^(2))`
`" "=10^(-5)sqrt(39.489 times 631.818)`
`" "=10^(-5)sqrt(671.307)`
`" "=10^(-5) times 25.91`
Resultant field at the centre = `2.591 times 10^(-4)T`
`" "alpha=tan^(-1)(B_(2)/B_(1))`
`" "=tan^(-1)((25.136 times 10^(-5))/(6.284 times 10^(-5)))`
`" "a=tan^(-1)(4) " w.r.t "B_(1)`
i.e., `" "a=75^(@)58^(') " w.r.t "B_(1)`
The direction of the resultant is `75^(@)58^(')` w.r.t. the direction of `B_(1)`