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(a) If a magnet of length l and magnetic moment M is bent in the form of a semicircular are then its new magnetic moment will be M' = 2M/π
(b) The magnetic moment of an electron due to its orbital motion is 1μB whereas that due to its spin motion it is μB/2.
i.e. Morbital =
and Mspin = s
Here μB = Bohr magneton
(i) The value of Bohr magneton μB = eh / 4πm
(ii) μB = 0.93 × 10–23 Amp-m2
(c) Other formulae of M:
(i) M = ni r2
(ii) M = eVr/2 = er2ω/2 = er22πf/2 = er2π/T
(iv) M = nωB
(d) Resultant magnetic moment :
(i) When two bar magnets are lying mutually perpendicular to each other, then
M = √M12 + M22 = √2mpl
(ii) When two coils, each of radius r and carrying current i, are lying concentrically with their planes at right angles to each other, then
M = √M12 + M22 = √2iπr2 if M1 = M2
Illustration 5: A square loop OABCO of side l carries a current i. It is placed as shown in figure. Find the magnetic moment of the loop.
Solution: Magnetic moment of the loop can be written as,
Illustration 6: Find the magnitude of magnetic moment of the current carrying loop ABCDEFA. Each side of the loop is 10 cm long and current in the loop is i = 2.0 A.
Solution : By assuming two equal and opposite currents in BE, two current carrying loop (ABEFA and BCDEB) are formed. Their magnetic moments are equal in magnitude but perpendicular to each other. Hence
Mnet = √M2 + M2 = √2 M
Where M = ia = (2.0)(0.1)(0.1) = 0.02 A-m2
Mnet = (√2) (0.02) A – m2 = 0.028 A-m2
Illustration 7: Two identical bar magnets each of length L and pole–strength m are placed at right angles to each other with the north pole of one touching the south pole of the other. Evaluate the magnetic moment of the system.
Solution: As magnetic moment is a vector
MR = (M12 + M22 + 2M1M2 cos θ)1/2 with tanΦ M2 sinθ / M1 + M2 cosθ and as here M1= M2 = mL and θ = 90°, so
MR = (M2 + M2 + 2MM cos 90)1/2 = (√2) mL
And, tanΦ Msin90 / M + Mcos 90 = 1, i.e., Φ = tan–1(1) = 45°