Tag: magnetic field due to bar magnet

Magnetic induction due to a short bar magnet on its axial line,

$B \displaystyle = \frac{\mu _0  2M}{4 \pi  d^3}$
At $\displaystyle \frac{d}{4} $ distance,
$B' \displaystyle = \frac{\mu _0 2M}{4 \pi (d/4)^3}$
$\displaystyle = \frac{\mu _0  2M}{4 \pi d^3} \times 64 = 64 B$

For a short Bar Magnet, the magnetic induction at a point on the axix at a distance $r$ from centre is given by  the formula

$r = \sqrt{13} d ;  2l  = 2d$
$B _eq = \displaystyle \frac{\mu _0}{4 \pi} \times \frac{\mu}{(r^2 + 1^2)^{\frac{3}{2}}}$
$\displaystyle =\frac{\mu _0}{4\pi} \times \frac{M}{\left ((\sqrt{13}d)^2 + (2d)^2 \right )^{\frac{3}{2}}}$
$=\displaystyle \frac{\mu _0}{4\pi} \times \frac{M}{(17 d^2)^{\frac{3}{2}}} = \frac{\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$

$B _{axial} = \displaystyle \frac{\mu _0}{4\pi} \frac{2Mr}{(r^2-l^2)^2}$
$B _{20} : B _{30} = 128  :  27$
$\displaystyle \frac{20}{(20^2 - l^2)^2} \times \frac{(30^2 - l^2)^2}{30} = \frac{128}{127}$
$2 (30^2 - l^2)^2 (27) = 3 (20^2 - l^2)^2 128$
$\sqrt{54}(900 - l^2) = \sqrt{384} (400 - l^2)$
$900 \sqrt{54} - \sqrt{54}l^2 = \sqrt{384} \times 400 - \sqrt{384}l^2$
$l^2 (12.2474) = 12224. 75$
$l^2 = 100 ;  l = \sqrt{100}-10 cm$
$\therefore$ Magnetic length $=2l = 20 cm$

$B _{equi} = \displaystyle \frac{\mu _0}{4 \pi } \frac{\mu}{r^3} = \frac{10^{-7}\times 5.4}{(0.3)^3} = 2 \times 10^{-5}T$

$B = \displaystyle \frac{\mu _0  2M}{4 \pi  d^3}$
At $(4d/5) $ distance
$B' = \displaystyle \frac{\mu _0  2M}{4 \pi \left ( \frac{4d}{5} \right )^3} = \frac{\mu _0  2M}{4  \pi (d^3)} \left ( \frac{125}{64} \right )$
$=\displaystyle \frac{125}{64}B$

When the north pole of short bar magnet is facing North pole of the earth, at the neutral point P, which is on equatorial line. 
$B _H = \dfrac {\mu _0 M}{4\pi d^3} = B _1$ ............(1)
When the magnet is rotated by $90^o$, the magnetic induction at P which is on axial line,
$B _H = \dfrac {\mu _0 2M}{4\pi d^3} = B _2$ ............(2)
Therefore, net magnetic induction at P is
$B _{net} = \sqrt {(B _1^2 + B _2^2)}$
$B _{net} = \sqrt {(1^2 + 2^2)}B _H = \sqrt 5 B _H$

$B=\frac{2\mu _0m}{4 \pi d^{3}}$

$B equatorial =\frac{\mu _0m}{4 \pi d^{3}}$

$d _1=d$
$d _2=2d$
Also,
$m _1=M$
$m _2=8M$

external flux of a surface is given by : E.ds.

since, the flux through the cube would be $E\times a2 = x$

therefore for a sphere,  the flux would be $E.\times Φ a2$

which is equal to $Φ$