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Magnetic Field Intensity | Intensity of Magnetic Field | Magnetic Field due to Bar Magnet

Magnetic Field Intensity | Intensity of Magnetic Field | Magnetic Field due to Bar Magnet :- The space around a bar magnet/current carrying conductor/current carrying solenoid in which magnetic effects can be experienced is called magnetic field. We know that if a charge particle (q) moves in a magnetic field B, then the force experienced by the charge particle is given by,

$\displaystyle \overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

If v = 1 m/s, q = 1 C and θ = 90° then, B = F, so,

“The Magnetic Field Intensity/strength of a magnetic field at a point can be defined as the force experienced by a unit positive charge particle moving with unit velocity in a direction perpendicular to the magnetic field.”

The Magnetic Field Strength is a vector quantity.

Unit of Magnetic Field Intensity

SI unit :-

From the above definition, F=qvB ⇒ B=F/qv

SI unit :-N/Am or Tesla

CGS unit :-

CGS unit of Magnetic Field Intensity is Gauss(G).

$\displaystyle 1G={{10}^{-4}}T$

Another definition :-

“Force experienced by unit north pole when it is placed in a magnetic field is called the magnetic field strength.”

Magnetic Field due to Bar Magnet on it’s axis

(Magnetic Field Intensity)

Let the pole strength/power of a bar magnet be m and its effective length be 2l. We have to calculate magnetic field strength at a point P located at a distance r from the center O on its axial position.

Magnetic field due to North Pole at point P :-

$\displaystyle {{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\frac{m}{{{(r-l)}^{2}}}$ …..(1) (away form north pole)

Magnetic field due to South Pole at point P :-

$\displaystyle {{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\frac{m}{{{(r+l)}^{2}}}$ …..(2) (towards north pole)

Net magnetic field at point P :-

B_axis = B₁ – B₂, (∵ B₁ > B₂)

$\displaystyle {{B}_{axis}}=\frac{{{\mu }_{0}}m}{4\pi }\left[ \frac{1}{{{(r-l)}^{2}}}-\frac{1}{{{(r+l)}^{2}}} \right]=\frac{{{\mu }_{0}}m}{4\pi }\left[ \frac{4rl}{{{({{r}^{2}}-{{l}^{2}})}^{2}}} \right]$

$\displaystyle \Rightarrow {{B}_{axis}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2\left[ m(2l) \right]r}{{{({{r}^{2}}-{{l}^{2}})}^{2}}}$

Here m (2l) = M (magnetic dipole moment of bar magnet)

so,

$\displaystyle {{B}_{axis}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2Mr}{{{({{r}^{2}}-{{l}^{2}})}^{2}}}$ …..(3)

Special situation :- if the bar magnet is small, then r >> l,

$\displaystyle \therefore {{B}_{axis}}\cong \frac{{{\mu }_{0}}}{4\pi }\frac{2M}{{{r}^{3}}}$ …..(4)

Magnetic Field due to Bar Magnet on it’s equatorial line

(Magnetic Field Intensity)

Magnetic field due to North Pole at point P :-

$\displaystyle {{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\frac{m}{{{\left( \sqrt{{{r}^{2}}+{{l}^{2}}} \right)}^{2}}}$ …..(5) (along NP)

Magnetic field due to South Pole at point P :-

$\displaystyle {{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\frac{m}{{{\left( \sqrt{{{r}^{2}}+{{l}^{2}}} \right)}^{2}}}$ …..(6) (along PS)

From equations (5) and (6),

$\displaystyle {{B}_{1}}={{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\frac{m}{{{r}^{2}}+{{l}^{2}}}=B$

Resultant magnetic field at point P :-

B_eq. = 2B cosθ

$\displaystyle \Rightarrow {{B}_{eq.}}=2\left( \frac{{{\mu }_{0}}}{4\pi }\frac{m}{{{r}^{2}}+{{l}^{2}}} \right)\left( \frac{l}{\sqrt{{{r}^{2}}+{{l}^{2}}}} \right)$