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. `vecmu_(2)=0`
B. `vecmu_(1)` and `vecmu_(2)` are in the same direction
C. `|vec(mu_(1))|/|vec(mu_(2))|=sqrt2`
D. `|vec(mu_(1))|/|vec(mu_(2))|=((1)/(sqrt2))`

Answer

C

Explanation

Initial magnetic moment `=mu_(1)=iL^(2)`

After folding the loop , `M=` magnetic moment due to
each part ` =i((L)/(2))xxL=(iL^(2))/(2)=(mu_(1))/(2)`
` implies mu_(2)=Msqrt2=(mu_(1))/(2)xxsqrt2=(mu_(1))/(sqrt2)`