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 = 7.5Am2

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 
B1=μ04π2M1r3Now Potential Energy of the system = U = -M.BSo Potential energy of M2 in the field of M1will beU = -M2.B1=-μ04π2M1M2r3As M2 is parallel to B1 i.e. θ =00Force =F=-dUdrForce on M2 due M1 will beF=-ddr-μ04π2M1M2r3F= -μ04π6M1M2r4F =-10-7×6x7.5x7.524F=-5.625x10-6T

If the similar poles of the two magnets face each other θ = 1800 and so U and F will be positive.
Regards.

 

  • -27
Hello Apoorva dear, the field due to bar magnet with pole strength 75 A m and length 0.1 m will be
10^-7 * 2 * 75 * (0.1) d / (d^2 - 0.1^2)^2
One pole is at a distance of 15 cm and the other pole is at a distance of 25 cm
So we have to find the forces on 75 Am one attractive and the other repulsive. 
With these two we have to find the resultant. Hope you would find it on your own. 
 
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