A steady current `i` flows in a small square lopp of wire of side `L` in a horizontal plane. The loop is now folded about its middle such that half of it lies in a vertical plane. Let `vecmu_(1)` and `vecmu_(2)` respectively denote the magnetic moments due to the current loop before and after folding. Then

Options

A. `vec(mu)_(2) = 0`
B. `vec(mu)_(1)` and `vec(mu)_(2)` are in the same direction
C. `(|vec(mu)_(1)|)/(|vec(mu)_(2)|) = sqrt(2)`
D. `(|vec(mu)_(1)|)/(|vec(mu)_(2)|) = ((1)/(sqrt(2)))`

Answer

C

Explanation

`vec(mu)_(1) = iL^(2) hat(k)`
`|vec(mu)_(1)| = iL^(2)`

`vec(mu)_(2) = i(L^(2))/(2)hat(i) + i(L^(2))/(2)hat(k)`
`|vec(mu)_(2)| = (iL^(2))/(2) xx sqrt(2) = (iL^(2))/(sqrt(2))`
`(|vec(mu)_(1)|)/(|vec(mu)_(2)|) = sqrt(2)`