A uniform wire of length L carrying a steady current I is first wound into a circular coil of N₁ turns, producing a magnetic dipole moment M₁ . The same wire is then unwound and rewound into another circular coil of N₂ turns, producing a magnetic dipole moment M₂ . If the ratio of their magnetic dipole moments M₁ M₂ is 4 , what is the ratio of their number of turns N₁ N₂ ?

Step-by-step solution

For a circular coil of N turns made from a wire of length L , the radius R is given by L = N(2 R) , which means R = L 2 N . The magnetic dipole moment is M = NIA = NI( R^2) . Substituting R , we get M = NI ( L 2 N )^2 = IL^2 4 N . This shows that for a fixed length of wire, the magnetic moment is inversely proportional to the number of turns ( M 1 N ). Given M₁ M₂ = 4 , we have N₂ N₁ = 4 , which gives N₁ N₂ = 1 4 . Answer: 1:4