Q.A short bar magnet has a magnetic moment of $0.48\ \text{J T}^{-1}$. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of $10\ \text{cm}$ from the centre of the magnet on

Concept and Intuition

A bar magnet behaves, at points far from it compared to its own length (the "short magnet" or point-dipole approximation), like a magnetic dipole. On its axis, the field is $B_{axial}=\dfrac{\mu_0}{4\pi}\dfrac{2m}{d^3}$ — twice as strong as on the equatorial line at the same distance, and falling off rapidly ($\propto 1/d^3$) because it is a dipole field, not a monopole field. This rapid fall-off is why bar-magnet fields become negligible just a short distance away.

Step-by-Step Solution

1. Axial field formula: $B = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2m}{d^3}$, with $\dfrac{\mu_0}{4\pi}=10^{-7}\ \mathrm{T\,m/A}$.
2. Substitute $m=0.48\ \mathrm{J\,T^{-1}}$, $d=0.1$ m: $d^3 = 0.001\ \mathrm{m^3}$.
3. $B = 10^{-7}\times\dfrac{2\times0.48}{0.001} = 10^{-7}\times960 = 9.6\times10^{-5}$ T. …

Concept and Intuition

Point P is end-on (axial) to magnet A, whose horizontal field is $B_1 = \dfrac{\mu_0}{4\pi}\dfrac{2M}{d_1^{3}}$. P is broadside (equatorial) to the perpendicular magnet B, whose field is $B_2 = \dfrac{\mu_0}{4\pi}\dfrac{M}{d_2^{3}}$ and points vertically. The needle aligns with the resultant, tilted at $\theta$ from the horizontal $B_1$ direction.

Step-by-Step Solution

1. Axial field of A at P: $B_1 = \dfrac{\mu_0}{4\pi}\dfrac{2M}{d_1^{3}}$ (horizontal).
2. Equatorial field of B at P: $B_2 = \dfrac{\mu_0}{4\pi}\dfrac{M}{d_2^{3}}$ (vertical).
3. The deflection satisfies $\tan\theta = \dfrac{B_2}{B_1} = \dfrac{M/d_2^{3}}{2M/d_1^{3}} = \dfrac{d_1^{3}}{2\,d_2^{3}}$.
4. Rearrange: $\dfrac{d_1^{3}}{d_2^{3}} = 2\tan\theta \Rightarrow \dfrac{d_1}{d_2} = (2\tan\theta)^{1/3}$.

Common Mistakes …

Concept and Intuition

A current loop behaves like a magnetic dipole, with field lines emerging from one face (like a bar magnet's north pole) and entering the opposite face (like the south pole). The right-hand rule — curl the fingers in the direction of current flow, and the thumb points in the direction of the magnetic field along the axis — identifies which face is the "north" face: the field lines point out of that face, exactly the way they emerge from a bar magnet's north pole.

Step-by-Step Solution

1. Curl the right-hand fingers along the direction of the current in the loop.
2. The thumb then points along the axis of the loop, in the direction of the magnetic field inside/through the loop.
3. The face from which the field lines emerge (the thumb's side) is, by analogy to a bar magnet, the north pole of this equivalent magnetic dipole. …

Concept and Intuition

In electrostatics, a single point charge (an "electric monopole") is the most basic and commonly observed object — charges of a single sign exist freely (e.g., an electron or proton). Electric dipoles (pairs of equal and opposite charges) also exist as composite structures. In magnetism, however, every magnet observed in nature has both a north and a south pole together; cutting a bar magnet in half only produces two smaller magnets, each with its own north-south pair, never an isolated pole. This experimental fact is formalized as Gauss's law for magnetism ($\oint\vec B\cdot d\vec A=0$), which states magnetic field lines have no beginning or end — unlike electric field lines, which start and end on individual (monopole) charges.

Step-by-Step Solution

1. Electric dipoles: composite systems of equal and opposite charges — these exist (e.g. polar molecules).
2. Electric monopoles: single isolated charges — these definitely exist (e.g. a free electron or proton).
3. Magnetic dipoles: bar magnets, current loops — these exist and are the most basic magnetic sources found. …

Concept and Intuition

Unlike electric field lines, which begin and end on charges (giving nonzero net flux, per Gauss's law of electrostatics), magnetic field lines always form closed loops with no beginning or end — there are no isolated magnetic "charges" (monopoles). Hence the net flux out of any closed surface for $\vec B$ is always exactly zero, regardless of what currents or magnets are inside or outside.

Step-by-Step Solution

1. Compare with Gauss's law for electricity: $\int \vec E\cdot d\vec s = q/\varepsilon_0$ (nonzero when charge is enclosed). …

Concept and Intuition

A bar magnet's poles arise from an unbroken chain of aligned atomic dipoles running along its length, N at one end and S at the other. Cutting it does not erase this alignment — each half is simply a shorter version of the same chain, so each half becomes a complete magnet on its own, with a north and a south pole. The end that was already a pole (say N) stays N. The newly exposed face at the cut must become the other pole (S), because inside a magnet field lines run from S to N — the cut face, which used to be the interior mid-point where the field pointed toward the N end, becomes the new S pole for the left piece. By the same logic the right piece's cut face becomes a new N pole.

Step-by-Step Solution

1. Original magnet: N (left end) ——— S (right end), dotted cut line at the midpoint.
2. After cutting, left piece: retains N at its left end; develops a new S pole at its cut (right) face.
3. Right piece: retains S at its right end; develops a new N pole at its cut (left) face.
4. The two pieces are held apart "as they are" — i.e., in their original relative alignment, so the left piece's cut face (S) is nearest the right piece's cut face (N). …

Concept and Intuition

"Neutral points" on the equatorial line of a bar magnet are where the magnet's own (equatorial) field exactly cancels the Earth's horizontal field, so their location pins down the magnet's moment. The equatorial field of a short bar magnet at distance $d$ from its centre is $B_{eq}=\dfrac{\mu_0}{4\pi}\dfrac{m}{(d^2+l^2)^{3/2}}$, where $m=q_m\times 2l$ is the magnetic moment and $l$ is the half-length.

Step-by-Step Solution

1. Half-length: $l = 16/2 = 8\,\mathrm{cm} = 0.08\,\mathrm{m}$.
2. Neutral points are symmetric about the centre, separated by $12\,\mathrm{cm}$, so each is at $d = 6\,\mathrm{cm} = 0.06\,\mathrm{m}$ from the centre.
3. At a neutral point: $B_{eq} = H \Rightarrow \dfrac{\mu_0}{4\pi}\cdot\dfrac{q_m(2l)}{(d^2+l^2)^{3/2}} = H$.
4. Compute $d^2+l^2 = 0.0036+0.0064 = 0.01\,\mathrm{m^2}$, so $(d^2+l^2)^{3/2} = (0.01)^{1.5} = 0.001\,\mathrm{m^3}$. …