Question

Derive an expression for magnitude of magnetic dipole moment
of a revolving electron.
A circular coil of 300 turns and diameter 14 cm carries a current
of 15A. Calculate the magnitude of the magnetic dipole moment
associated with the coil.​

Answer 1

The magnitude(M) of the magnetic dipole moment = 69.27 Am^2

Explanation:

Assuming the angular velocity with which the electron rotates about an axis be w,

so,

Angular velocity = w rod/s

w = 2πf

f = (w/2π)Hz

As we know,

Rotating charge follows generated current,

Equivalent current = qf

= wq/2π

The given case can be considered as a case of a current-bearing loop.

Assuming the radius be r, so

vector r = πr^2n

n cap - direction is ⊥ to the rotation,

Magnetic moment = IA

= wq/2π * πr^2

= wqr^2/2

In favor of an electron,

q = e = 1.6×10^-19 C

∴ Magnetic moment = wer^2/2

Given,

N = 300,

D = 14cm

r = 14/2 = 7 × 10^-2m

(current) i = 15 A

Magnetic dipole moment associated with the coil,

M = NI(πr^2)

⇒ M = 300 * 15 * 22/7 * (7 × 10^-2)^2

∵ M = 69.27 Am^2

Learn more: Magnitude

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