Manipal Entrance Test 2015 Physics Paper

Magnetic moment of a loop is given by pm=IA
where, I = current through the loop
A = area of loop
Perimeter of square should be equal to the circumference of circle, since wires have same length.
Therefore, 4a=2πr⇒a=

πr

2

...(i)
In Fig. (i), magnetic moment of square loop is given by
pm1=I(A1)=

Iπ2r2

4

[from Eq. (i)] ...(ii)
In Fig. (ii), magnetic moment of circular loop is given by
pm2=I(A2) =I(πr2) ...(iii)
From Eqs. (ii) and (iii) we get
⇒

pm1

pm2

=

π

4