The ratio of magnetic field at the centre of the current carrying circular loop and magnetic moment is ' x '. When both the current and radius are double then the ratio will be

Step-by-step solution

The magnetic field at the center of the wire is given as B= ₀ i 2 r . The magnetic moment is given as M = i r ^2 The ratio is ₀ i 2 r i r ^2 =x ....(i) x= ₀ 2 ^2 r^3 When current and radius is doubled, x₁= ₀ 2 i 2 2 r 2 i (2 r)^2 ....(From (i)) x₁= ₀ 16 ^2 r^3 x ₁= 1 8 ( ₀ 2 ^2 r ^3 ) x₁= x 8