Unit 03: Magnetism and Matter • Chapter 05
Bar Magnet as an Equivalent Solenoid & Field Lines
Continuous closed magnetic loops, solenoid slice integration, dipole analogies, and far-field inverse-cube calculations.
Socho ek lambe solenoid ko jisme hazaron circular current rings ek ke baad ek saji hain. Har ek ring ek micro-magnetic dipole banati hai. Jab aap in sabhi rings ko combine karte hain, toh andar adjacent faces ke currents aapas mein cancel ho jaate hain aur sirf outer surface par ek helical sheet of current bachti hai. Theek yahi cheez ek Permanent Bar Magnet ke andar hoti hai: atoms ke andar ghoomte electrons ke microscopic current loops add hokar ek macroscopic magnetic dipole moment banate hain. Isliye chahe aap iron filings ko solenoid ke paas chhidkein ya bar magnet ke paas, unka magnetic field pattern 100% identical banta hai!
1. Magnetic Field Lines Axioms
Universal rules governing magnetic vector trajectories:
- Form continuous closed loops (outside from N to S, inside from S to N).
- Tangent at any point gives direction of net magnetic field $\vec{B}$.
- Two lines never intersect (otherwise two field directions would exist).
- Crowding of lines represents relative magnitude of field strength.
2. Dipole Field Comparison ($r \gg a$)
Exact parallel with Electrostatic Dipole ($p \to M, \frac{1}{4\pi\varepsilon_0} \to \frac{\mu_0}{4\pi}$):
Notice: $B_{\text{axial}} = 2 \times B_{\text{equatorial}}$ for the same distance $r$.
Axial Magnetic Field of a Finite Solenoid as a Bar Magnet
Fig 5.1A: Elemental slice $dx$ of a finite solenoid of length $2l$ and radius $a$ evaluated at axial point $P$.
1. Consider a solenoid of length $2l$, radius $a$, carrying current $I$ with $n$ turns per unit length. Total turns $N = n(2l)$.
2. Take an elemental circular slice of thickness $dx$ at distance $x$ from the center $O$. Number of turns in this slice is $n\,dx$.
3. Distance of point $P$ from the slice is $(r - x)$. Using the circular loop axial field formula:
4. For a distant axial point ($r \gg a$ and $r \gg x$), we can approximate $(r - x) \approx r$ and neglect $a^2$ in the denominator: $[(r - x)^2 + a^2]^{3/2} \approx r^3$:
5. Integrate over the entire length of the solenoid from $x = -l$ to $x = +l$:
6. Multiply and divide numerator by $\pi$:
7. Since magnetic dipole moment $M = N I A = N I (\pi a^2)$, multiplying numerator and denominator by 2:
Deduction: This expression is completely identical to the electric field of an electric dipole along its axis ($E = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}$), proving that a finite solenoid acts as an exact equivalent bar magnet.
Trap 1: Electric Field Lines vs Magnetic Field Lines Continuity
Electrostatic field lines are discontinuous—they start at positive charges and end at negative charges (cannot exist inside a conductor). In contrast, magnetic field lines form continuous closed loops everywhere (outside North to South, inside South to North). This is because isolated magnetic monopoles do not exist ($\oint \vec{B}\cdot d\vec{A} = 0$).
Trap 2: Cutting a Bar Magnet (Length vs Cross-Section)
Let original dipole moment be $M = m \times 2l$, where $m$ is pole strength and $2l$ is length:
• Cut transverse to length (into 2 halves): Length becomes $l$, pole strength $m$ is unchanged $\implies \mathbf{M' = M/2}$.
• Cut along length (longitudinally): Length remains $2l$, pole strength halves to $m/2 \implies \mathbf{M' = M/2}$.
In both cases, each piece becomes a complete new dipole with both N and S poles!
Trap 3: The Factor of 2 in Axial vs Equatorial Ratio
At equal large distance $r$ ($r \gg l$): $$B_{\text{axial}} = 2 \times B_{\text{equatorial}}$$ Direction Alert: $\vec{B}_{\text{axial}}$ points parallel to $\vec{M}$ (from S to N), whereas $\vec{B}_{\text{equatorial}}$ points antiparallel to $\vec{M}$ (from N to S). Vector equation: $\vec{B}_{\text{eq}} = -\frac{\mu_0}{4\pi}\frac{\vec{M}}{r^3}$.
Trap 4: Magnetic Length vs Geometric Length
Poles of a bar magnet reside slightly inside the physical edges. The magnetic length ($2l$) between poles is approximately 84% of its geometric length: $$L_{\text{magnetic}} \approx 0.84 \times L_{\text{geometric}}$$
Problem: A short bar magnet placed with its axis at $30^\circ$ with an external field of $800\text{ G}$ experiences a torque of $0.016\text{ N}\cdot\text{m}$. (a) What is the magnetic moment of the magnet? (b) What is the magnitude of the magnetic field at a point on its axis at a distance of $10\text{ cm}$ from its center? (Take $1\text{ G} = 10^{-4}\text{ T}$).
Solution:
1. External field $B = 800\text{ G} = 800 \times 10^{-4}\text{ T} = 0.08\text{ T}$, $\theta = 30^\circ$, torque $\tau = 0.016\text{ N}\cdot\text{m}$.
Using $\tau = M B \sin\theta$:
2. Axial field at distance $r = 10\text{ cm} = 0.1\text{ m}$:
Problem: A thin bar magnet of length $L$ and magnetic moment $M$ is bent into a semicircle. What will be the new magnetic dipole moment of the bent magnet?
Solution:
1. Let $m$ be the pole strength of the magnet. Initial magnetic moment: $M = m \cdot L$.
2. When bent into a semicircle of radius $R$, the circumference equals length $L$:
3. The new distance between the two poles (diameter of the semicircle) is $2R$:
4. Since bending does not change the pole strength $m$, the new magnetic moment $M'$ is:
Trap Alert: Pole strength $m$ is an intrinsic cross-sectional property; it does not change on bending, but effective pole displacement shrinks from $L$ to $2L/\pi$.
Axial Field
$B = \frac{\mu_0 2M}{4\pi r^3}$
Equatorial Field
$B = \frac{\mu_0 M}{4\pi r^3}$
Dipole Torque
$\vec{\tau} = \vec{M}\times\vec{B}$
Gauss Law Magnetism
$\oint \vec{B}\cdot d\vec{A} = 0$