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Two bar magnets of length 0.1m and pole strength 75 Am each are placed on the same line.The distance between their centres is .2m.What is the resultant force due to one on the other when

(1)the north pole of one faces the south pole of the other

(2)the north pole of one faces the north pole of the other

Dear Student,

Magnetic Moment = M = Pole Strength x Length of the Magnet

M = mL = 75Am x 0.1m = 7.5Am^{2}

If the opposite poles of two magnets face each other the field due to M1 (magnet -1)at the position of M2 (magnet -2)will be

$\underset{B1}{\to}=\frac{{\mu}_{0}}{4\mathrm{\pi}}\frac{2M1}{{r}^{3}}\phantom{\rule{0ex}{0ex}}NowPotentialEnergyofthesystem=U=-\underset{M}{\to}.\underset{B}{\to}\phantom{\rule{0ex}{0ex}}SoPotentialenergyof\underset{M2}{\to}inthefieldof\underset{M1}{\to}willbe\phantom{\rule{0ex}{0ex}}U=-\underset{M2}{\to}.\underset{B1}{\to}=-\frac{{\mu}_{0}}{4\mathrm{\pi}}\frac{2M1M2}{{r}^{3}}\phantom{\rule{0ex}{0ex}}AsM2isparalleltoB1i.e.\theta ={0}^{0}\phantom{\rule{0ex}{0ex}}Force=F=-\frac{dU}{dr}\phantom{\rule{0ex}{0ex}}Forceon\underset{M2}{\to}due\underset{M1}{\to}willbe\phantom{\rule{0ex}{0ex}}F=-\frac{d}{dr}\left[-\frac{{\mu}_{0}}{4\mathrm{\pi}}\frac{2M1M2}{{r}^{3}}\right]\phantom{\rule{0ex}{0ex}}F=-\frac{{\mu}_{0}}{4\mathrm{\pi}}\frac{6M1M2}{{r}^{4}}\phantom{\rule{0ex}{0ex}}F=-{10}^{-7}\times \frac{6x7.5x7.5}{{2}^{4}}\phantom{\rule{0ex}{0ex}}F=-5.625x{10}^{-6}T$

If the similar poles of the two magnets face each other θ = 180^{0 }and so U and F will be positive.

Regards.

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