Question

A current enters a uniform conducting loop of radius r at X and leaves at Y
(i) What is magnetic field at O due to arc portions XBY and XAY of the loop ?
(ii) What is net magnetic field at O ?

Answer 1

Let us first find out currents in the arc portions, which are in parallel across points X and Y
`rArr (i_(XAY))/(i_(XBY)) = (R_(XBY))/(R_(XAY)) = (l_(XBY))/(l_(XAY)) = (theta )/(2pi - theta)`
(Here i, R and l represent current, resistance and length respectively)
Therefore,
`i_(XAY) = (itheta)/(2pi), i_(XBY) = (i(2pi - theta))/(2pi)`
`rArr B_(XAY)=` Magnetic field due to arc portion XAY
`= (mu_(0) ((itheta)/(2pi)) (2pi - theta))/(4pi r) ox = (mu_(0) i theta (2pi - theta))/(8pi^(2) r) ox`
and `B_(XBY) = (mu_(0)[(i(2pi - theta))/(2pi)] theta)/(4pi r) o. = (mu_(0) itheta(2pi - theta))/(8pi^(2) r) o.`
Net Magnetic field `= vecB_(XAY)^(•) + vec(B)_(XBY) = 0` at O