Physics · Magnetism And Matter · NEET
Magnetic moment is M = m x L, where m is pole strength and L is the distance between the two poles. Bending does not change the pole strength m (that depends on the material and cross-section). But bending brings the two end poles closer in a straight line, so the effective distance d shrinks. Since M = m x d, a smaller d means a smaller moment. The mass and pole strength are unchanged; only the geometry (pole separation) changed.
Use the straight-line distance between the two free poles. Do NOT vector-add the two arm moments. The reason: when you bend at the middle, the middle point now has an N pole from one half and an S pole from the other half sitting together, and they cancel. Only the two outer free poles matter. So M_new = m x d, where d is the straight distance between those two outer poles. The vector-sum shortcut gives a wrong answer here.
First find m from the ORIGINAL magnet: m = M / L, where M is the given moment and L is the original full length. Bending does not change m. Then compute the new pole separation d from the geometry of the bend, and finally M_new = m x d = (M/L) x d.
Each arm has length L/2 (bent at the middle). The two arms and the line joining the free poles form a triangle with two equal sides of L/2 and an included angle of 60 degrees. A triangle with two equal sides and a 60-degree angle between them is equilateral, so the third side (the pole separation) also equals L/2. Then M_new = (M/L) x (L/2) = M/2.
Each arm is L/2. With a 90-degree angle between them, the pole separation is the hypotenuse: d = sqrt((L/2)^2 + (L/2)^2) = L/sqrt(2). So M_new = (M/L) x (L/sqrt2) = M/sqrt2. This is a common variation NTA can ask, so learn the method, not just the 60-degree answer.
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 60 degrees with each other. The magnetic moment of this new magnet is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
M_new = m x d, where m = M/L is the (unchanged) pole strength and d is the new straight-line distance between the two free end poles after bending.
No. Pole strength depends on the material and the cross-sectional area, neither of which changes when you bend it. Only the pole separation changes.
For a straight magnet of length L and moment M, m = M/L. Bent into a semicircle of radius r, the length becomes the arc: L = pi x r, so r = L/pi. The two ends (poles) are at the diameter, so d = 2r = 2L/pi. Then M_new = (M/L)(2L/pi) = 2M/pi.
Because bending always brings the two poles closer in a straight line (d becomes less than L). Since M = m x d and m is fixed, a smaller d always gives a smaller moment. The only case where the moment stays M is a straight (180-degree) magnet.
No. This question is only about the magnitude of the magnetic moment vector after bending. Torque (tau = MB sin theta) and potential energy (U = -MB cos theta) use this moment M as an input, so you often find the bent moment first, then plug it into those formulas.
