Question

An iron bar of length L has magnetic moment M. It is bent at the middle of its length such that the two arms make an angle 45° with each other. The magnetic moment of this new magnet is :

• A. M
• B. M/2
• C. M/sqrt(2)
• D. M(sqrt(2) + 1)/2

Answer 1

Answer: (C)

M/sqrt(2)

Answer 2

Let the pole strength of the magnet be m. The magnetic moment M = m * L When the magnet is bent, the new length is given by L' = 2 * (L/2) * sin(theta/2) = L * sin(theta/2) Here theta = 45 degrees So L' = L * sin(45/2) = L * sin(22.5) The new magnetic moment M' = m * L' = m * L * sin(22.5) However, a simpler approach is: M' = M cos(theta/2) where theta is the angle between the two arms. M' = M cos(45/2) = M cos(22.5) The effective length is halved, so the pole strength remains the same. The new magnetic moment is M' = m * L' L' can also be calculated using the cosine rule: L'^2 = (L/2)^2 + (L/2)^2 - 2(L/2)(L/2)cos(180-45) L'^2 = L^2/4 + L^2/4 - (L^2/2)cos(135) L'^2 = L^2/2 - (L^2/2)(-1/sqrt(2)) L'^2 = L^2/2 + L^2/(2sqrt(2)) L' = L * sqrt(1/2 + 1/(2sqrt(2))) A simpler approach: The new magnetic moment M' = m * effective length Effective length = sqrt((L/2)^2 + (L/2)^2) = sqrt(L^2/4 + L^2/4) = sqrt(L^2/2) = L/sqrt(2) Since the pole strength remains the same, M' = m * L/sqrt(2) = M/sqrt(2)