Magnetism Concept Page - 2

Diagram

Neutral points in bar magnet

The diagram shows the locations of neutral points of bar magnet on earth surface in two configurations of magnetic fields.

Result

Magnetic lines of bar magnet

Magnetic lines of force of a magnet shows the direction in which a magnetic material will be aligned when placed at the location. Magnetic field lines for a bar magnet are as shown in the diagram. 

Definition

Properties - Magnetic lines of force

Properties of magnetic lines of force are:

1. They are continuous closed curves.
2. They originate from the North Pole and terminate at the South Pole.
3. Two lines do not intersect each other.
4. They are crowded where the magnetic field is strong, i.e. near the poles.

Definition

Properties of magnetic fields of lines

Magnetic field lines are imaginary lines that represent magnetic fields. It gives direction of magnetic force. An advantage of using magnetic field lines as a representation is that laws of magnetism and (electromagnetism) can be stated using concepts like 'number' of field lines through a surface.

Magnetic field lines are like streamlines in fluid flow, they represent something continuous, and a different resolution would show more or fewer lines. Various phenomena have the effect of displaying magnetic field lines as though the field lines were physical phenomena. For example, iron filings placed in a magnetic field, form lines that correspond to field lines. Field lines can be used as a qualitative tool to visualize magnetic forces.' Unlike' poles of magnets attract because they are linked by many field lines, 'like' poles repel because their field lines do not meet, but run parallel, pushing on each other.

Example

Example - Magnetic filed strength due to a bar magnet

A magnet is 10$10$ cm long and its pole strength is 120$120$ CGS units (1$1$ CGS unit of pole strength=0.1 Am$0.1Am$). Find the magnitude of the magnetic field B$B$ at a point on its axis at a distance 20 cm from it.
Solution:
The pole strength is m=120$m=120$ CGS units = 12 Am$12Am$.
Magnetic field is 2l=10 cm$2l=10cm$ or l=0.05 m$l=0.05m$.
Distance from the magnet is d=20 cm=0.2 m$d=20cm=0.2m$.
The field B$B$ at a point in end-on position is:
B=μ04π2Md(d2−l2)2=μ04π2mld(d2−l2)2=(10−7TmA)4×(12Am)×(0.05m)×(0.2m)[(0.2m)2−(0.05m)2]2=3.4×10−5T$B={\displaystyle \frac{{\mu }_0}{4\pi }}{\displaystyle \frac{2Md}{{(d^2-l^2)}^2}}={\displaystyle \frac{{\mu }_0}{4\pi }}{\displaystyle \frac{2mld}{{(d^2-l^2)}^2}}=({10}^{-7}{\displaystyle \frac{Tm}{A}}){\displaystyle \frac{4\times (12Am)\times (0.05m)\times (0.2m)}{{[(0.2m{)}^2-(0.05m{)}^2]}^2}}=3.4\times {10}^{-5}T$

Definition

Magnetic Field strength due to bar magnet

Consider a bar magnet of length 2l and pole strength m. Suppose a point P on the axis of the magnet at a distance r from its center. (r-l) is the distance of P form the N-pole of the magnet. The magnetic field intensity at P due to the north-pole of the magnet is
B1=μom4πr2$B_1={\displaystyle \frac{{\mu }_{o}m}{4\pi r^2}}$B1=μom4π(r−l)2$B_1={\displaystyle \frac{{\mu }_{o}m}{4\pi (r-l{)}^2}}$
which is directly away from N-pole. Since the south of the magnet is at a distance  r + l from P, so magnetic field intensity at P due to S-pole isB2=μom4π(r+l)2$B_2={\displaystyle \frac{{\mu }_{o}m}{4\pi (r+l{)}^2}}$
which is direct towards, the S-pole of the magnet. The magnetic field intensity B at P is the resultant of these two fields                               B=B1+(−B2)$B=B_1+(-B_2)$
                               B=B1−B2)$B=B_1-B_2)$
                                   =μom4π×[1(r−l)2−1(r+l)2]$={\displaystyle \frac{{\mu }_{o}m}{4\pi }}\times [{\displaystyle \frac{1}{(r-l{)}^2}}-{\displaystyle \frac{1}{(r+l{)}^2}}]$ 
                                   =μom4π×4lr(r2−l2)2$={\displaystyle \frac{{\mu }_{o}m}{4\pi }}\times {\displaystyle \frac{4lr}{(r^2-l^2{)}^2}}$
                                   =μo2Md4π(r2−l2)2$={\displaystyle \frac{{\mu }_{o}2Md}{4\pi (r^2-l^2{)}^2}}$

Definition

Magnetic field follows superposition principle

The magnetic field of several sources is the vector addition of magnetic field of each individual source. Thus it follows the superposition principle.

Definition

The net magnetic flux through a closed surface is zero

The contribution to magnetic flux for a given area is equal to the area times the component of magnetic field perpendicular to the area.
For a closed surface the number of magnetic field lines entering the surface is equal to the number of field lines exiting it.Thus the sum of magnetic flux is always equal to zero.

Example

Coulomb's Law of magnetism for force between magnetic poles

Two equal poles each repel with a force of 10 newtons when placed 2 cm apart in air.Find the pole-strength of each in amp-meter.Use the formula:
F=μ04πQ1Q2r2$F={\displaystyle \frac{{\mu }_0}{4\pi }}{\displaystyle \frac{Q_1{Q}_2}{r^2}}$
∴10−7×Q2/(0.02)2=10$\therefore {10}^{-7}\times Q^2/(0.02{)}^2=10$
⇒m2=40000$\Rightarrow m^2=40000$
Q=200 Am$Q=200Am$

Definition

Calculate magnetic induction at a point along the axis of a bar magnet

Let BS$B_S$ and BN$B_N$ be the magnetic induction at south pole and north pole of a magnet respectively at a distance r$r$ from the magnet. Let the moment of south side is −m$-m$ and moment of north side is +m$+m$.
Let us consider a magnet of length 2d$2d$. 
Now BP=BN−BS$B_P=B_N-B_S$
BP=μ04π.m(π−d)2−μ04π.m(π−d)2$B_P={\displaystyle \frac{{\mu }_0}{4\pi }}.{\displaystyle \frac{m}{{(\pi -d)}^2}}-{\displaystyle \frac{{\mu }_0}{4\pi }}.{\displaystyle \frac{m}{{(\pi -d)}^2}}$
BP=μ0m4π.2μr(r2−d2)2$B_P={\displaystyle \frac{{\mu }_{0}m}{4\pi }}.{\displaystyle \frac{2\mu r}{{(r^2-d^2)}^2}}$ where μ=2md$\mu =2md$ which can be represnted as magnetic dipole M$M$.

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