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The Bar Magnet - Bar Magnet as an Equivalent Solenoid

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Estimated time: 8 minutes
CBSE: Class 12

Definition: Solenoid

A solenoid is a long cylindrical coil of insulated wire wound closely in the shape of a helix. When an electric current is passed through it, it produces a magnetic field similar to that of a bar magnet, with one end acting as a north pole and the other as a south pole.

CBSE: Class 12

Similarity Between a Bar Magnet and a Solenoid

Think of a bar magnet as being made up of an enormous stack of tiny current loops (like the rings of a solenoid), each carrying a small circulating current. This is Ampere's hypothesis in action. Just as a solenoid with many turns of current-carrying wire produces a magnetic field at its poles, the atomic circulating currents inside a bar magnet produce the same effect.

Analogy: Imagine stacking hundreds of identical ring-shaped magnets (each = one current loop of a solenoid) end-to-end. The resulting field pattern outside the stack is indistinguishable from that of a bar magnet. A solenoid carrying current is exactly this stack.

CBSE: Class 12

Derivation

Consider a solenoid with:

Symbol Meaning
a Radius of solenoid
2l Total length of solenoid (centre at O)
n Number of turns per unit length
I Current through solenoid
r Distance of observation point P on the axis
dx Small element at distance x from O

Step 1: Field due to one small element:

  • A small element of thickness dx at distance x from O has ndx turns. The field at axial point P (at distance r) due to this element is:
    dB = \[\frac{\mu_0ndxIa^2}{2\left[(r-x)^2+a^2\right]^{3/2}}\]

Step 2: Integrate over the full length:

  • Integrate dB from x = −l to x = +l:
    B = \[\frac{\mu_0nIa^2}{2}\int_{-l}^{+l}\frac{dx}{\left[(r-x)^2+a^2\right]^{3/2}}\]

Step 3: Apply the Far-Field Approximation:

  • For the far axial field, apply the conditions r ≫ a and r ≫ l. The denominator simplifies to r3:
    [(r − x)² + a²]³/² ≈ r3

Step 4: Final Result:

  • B = \[{\frac{\mu_0}{4\pi}\cdot\frac{2m}{r^3}}\]
    where m = nI(πa2)(2l) = total magnetic moment of the solenoid (= total turns × current × area).

Step 5: Conclusion:

  • This expression is identical to the far axial field of a bar magnet:
    Baxial (bar magnet) = \[{\frac{\mu_0}{4\pi}\cdot\frac{2m}{r^3}}\]

∴ The magnetic moment of the bar magnet = magnetic moment of the equivalent solenoid.

CBSE: Class 12

Key Points: Bar Magnet as an Equivalent Solenoid

  • A bar magnet can be thought of as a large number of atomic circulating currents (Ampere's Hypothesis).
  • The far axial magnetic field of both a bar magnet and a solenoid is B = \[\frac {μ_0}{4π}\]\[\frac {2m}{r^3}\]​, confirming equivalence.
  • Equivalence holds only at far axial distances (r ≫ a, r ≫ l).
  • The magnetic moment of the bar magnet = the magnetic moment of the equivalent solenoid (m = NIA).
  • Similarities: field pattern, axial field formula, poles, directive and attractive properties.
  • Key differences: a bar magnet is permanent, a solenoid is temporary; solenoid poles are reversible, bar magnet poles are fixed.
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