A current carrying loop consists of 3 identical quarter circles of radius R, lying in the positive quadrants of the x-y, y-z and z-x planes with their centres at the origin, joined together. Find the direction and magnitude to `vecB` at the origin.

Explanation

The magnetic field at the centre of a current carrying loop of radius R and carrying a current I is given by:
`B = (mu_(0)I)/(2 R) `
let us assume that the current I is carried by each quarter circles of radius R.
The magnitude of magnetic field produced by each quarter circle at its centre will be
`B_(1) = 1/4 ((mu_(0)I)/(2R))`
The magnetic field at the centre (origin) due to quarter circle lying in the x-y plane will be along Z-axis.
` rArr" " B_(xy) = B_(1) hat k`
` = 1/4 ((mu_(0)I)/(2R)) hat k`
Similarly, the magnetic field at the centre due to quarter circles lying in the y-z and z-x planes will be along X-axis and Y-axis respectively.
`rArr " " B_(yz) = B_(1) hat i `
` = 1/4 ((mu_(0)I)/(2R)) hat i `
` B_(zx) = B_(1) hat j`
` = 1/4 ((mu_(0)I)/(2R)) hat j `
The total magnetic at the centre ( origin) will be
` vec B = vecB_(xy) + vecB_(yz) + vecB_(zx)`
` = B_(1) hat k + B_(1) hati + B_(1) hat j`
` = B_(1) (hatk + hat i + hat j)`
` = 1/4 ((mu_(0)I)/(2R)) [ hatk + hat i + hat j]`
` = (mu_(0)I)/(8R) ( hat i + hatj + hatk)`
` :. ` Magnitude of `vecB` will be given by
`(mu_(0)I)/(8R) sqrt(1^(2) + 1^(2) + 1^(2)) = (sqrt3 mu_(0)I)/(8R)`