Question

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 0.2m. What is the resultant force due to one on the other when
(i) the north pole of one faces the south pole of the other and
(ii) the north pole of one faces the north pole of the other?

[S.L. Arora, Part-1, Class-XII, Page 5.5 , Q.no 5]

PLEASE ANSWER​

Answer 1

Answer:

The answer will be −5.625x10−6T

Explanation:

According to the problem the length of the magnets,l are 0.1 m

and the strength of the pole, m is 75 Am

Therefore the  Magnetic Moment p = Pole Strength x Length of the Magnet

p = ml = 75Am x 0.1m = 7.5 Am^2

let the magnetic field be B

and the two magnets let a1 and a2

Now, if the opposite poles of two magnets face each other the magnetic field due to a1 at the position of a2 will be  

B = μ0/4π  2a1/r^3

Let consider the potential energy of the system is E

E = -p.B

So Potential energy of a2 in the field of a1will be

E = −a2.B

  =−μ0/4π 2a1a2/r^3

Force =F

F=−dE/dr

Force on a2 due to a1 will be

F=−d/dr[−μ0/4π 2a1a2/r^3]

F= −μ0/4π 6a1a2/r^4

F =−10^(−7)×6x7.5x7.5/2^4

F=−5.625x10^(−6)T

Answer 2

Answer:

The above answer is correct.