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Magnetic Field of a Magnetic Dipole (Small Bar Magnet)

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Estimated time: 8 minutes
CISCE: Class 12

Introduction

A bar magnet has two magnetic poles: North (N) and South (S). When the separation between the poles is very small compared with the distance of the observation point from the magnet, it behaves as a short magnetic dipole.

A magnetic dipole produces a magnetic field that decreases rapidly with distance:

B ∝ \[\frac {1}{r^3}\]
Important: The magnetic field of a short bar magnet is similar in form to the electric field of an electric dipole. In both cases, the field varies as 1/r3.
CISCE: Class 12

Formula: Magnetic Dipole Moment

M = mp​(2l)

  • SI unit: A m2
  • Direction: from the South pole to the North pole.
CISCE: Class 12

Field on the Axial Line

Consider a bar magnet of pole strength mp, magnetic length 2l, and magnetic dipole moment M. Let point P lie on the axial line at distance r from its centre.

Exact Expression

The magnetic field at P is:

  • Baxial = \[\frac{\mu_0}{4\pi}\frac{2Mr}{(r^2-l^2)^2}\]

For a Short Bar Magnet

  • For r ≫ l: (r2 − l2)2 ≈ r4
  • Therefore, Baxial = \[{\frac{\mu_0}{4\pi}\frac{2M}{r^3}}\]

Direction: \[\vec B_{axial}\] is along \[\vec M\], i.e. from S to N.

CISCE: Class 12

Field on the Equatorial Line

Let point P lie on the equatorial line at a distance rr from the centre of the bar magnet.

Exact Expression

  • Bequatorial = \[\frac{\mu_0}{4\pi}\frac{M}{(r^2+l^2)^{3/2}}\]

For a Short Bar Magnet

  • For r ≫ l: (r2 + l2)3/2 ≈ r3
  • Hence, Bequatorial = \[{\frac{\mu_0}{4\pi}\frac{M}{r^3}}\]

Direction: \[\vec B_{equatorial}\] is opposite to \[\vec M\], i.e. from N to S.

CISCE: Class 12

Example

A short bar magnet has magnetic dipole moment M = 0.48 A m2. Find the magnetic field at a point 10 cm from its centre:

  1. on its axial line;
  2. on its equatorial line.

Take \[\frac {μ_0}{4π}\] = 10−7 T m A−1.

Given: M = 0.48 A m2 ,r = 10 cm = 0.10 m

(A) Axial Field

  • Ba = \[\frac{\mu_0}{4\pi}\frac{2M}{r^3}\]
  • Ba = \[10^{-7}\times\frac{2(0.48)}{(0.10)^3}\]
  • Ba = 9.6 × 10−5 T
  • Ba = 0.96 G

Direction: S to N along the magnetic axis.

(B) Equatorial Field

  • Be = \[\frac{\mu_0}{4\pi}\frac{M}{r^3}\]
  • Be = \[10^{-7}\times\frac{0.48}{(0.10)^3}\]
  • Be = 4.8 × 10−5 T
  • Be = 0.48 G

Direction: N to S, opposite to \[\vec M\].

Check: Ba = 2Be. Hence, the answer is consistent.

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