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Question: A small, magnetized sphere A of pole strength m is counter-poised by a pan in a balance. Another mag...

A small, magnetized sphere A of pole strength m is counter-poised by a pan in a balance. Another magnetised sphere B of the same mass as A is placed below A so that their centres are at a distance .If A and B are unlike poles of same strength and to restore the counter weight in the pan is 1cm1cm increased by 500gm500gm, the pole strength of each sphere is:
A) 1.4Am1.4Am
B) 9.8Am9.8Am
C) 0.7Am0.7Am
D) 70Am70Am

Explanation

Solution

The pole strength of a magnet is defined as the strength of a magnetic pole to attract a magnetic field towards itself is known as pole strength. It is a scalar quantity. It is the force exerted by one magnet on the other magnet. The poles have equal and opposite magnitudes.

Complete step by step solution:
Step I: Since the spheres are placed close to each other, so let their pole strengths be P1{P_1} and P2{P_2}. Since their pole strengths are equal so
P1=P2=P{P_1} = {P_2} = P
Also the distance between the two spheres is d=1cm=0.01md = 1cm = 0.01m

Step II: If the two poles are small then they can be represented as point magnetic charges. The force between two magnetic poles is given by the formula
F=μq1q24πr2F = \dfrac{{\mu {q_1}{q_2}}}{{4\pi {r^2}}}
Where FF is the force
q1,q2{q_1},{q_2}are magnitude of charges, in this case it is P1,P2{P_1},{P_2}
μ\mu is the permeability
rr is the distance

Step III:
F1=μP1P24πd2{F_1} = \dfrac{{\mu {P_1}{P_2}}}{{4\pi {d^2}}}
Since the pole strengths are equal, so
F1=μP24πd2{F_1} = \dfrac{{\mu {P^2}}}{{4\pi {d^2}}}---(i)

Step IV: When the weight 500gm500gm is added, then another force will act.
F2=mg{F_2} = mg
m=0.5kg;g=10m/s2m = 0.5kg;g = 10m/{s^2}
F2=0.5×10{F_2} = 0.5 \times 10---(ii)

Step V: Since the forces are balanced, so
F1=F2{F_1} = {F_2}
μP24πd2=0.5×10\dfrac{{\mu {P^2}}}{{4\pi {d^2}}} = 0.5 \times 10
Since the term 4π4\pi , so substituting all the other values and solving
107P2(0.01)2=0.5×10\dfrac{{{{10}^{ - 7}}{P^2}}}{{{{(0.01)}^2}}} = 0.5 \times 10
P2=0.5×10×0.0001107{P^2} = \dfrac{{0.5 \times 10 \times 0.0001}}{{{{10}^{ - 7}}}}
P2=4998.49{P^2} = 4998.49
P=70.7AmP = 70.7Am

Step VI: The pole strength of each sphere is 70.7Am.70.7Am.

Therefore, Option (D) is the right answer.

Note: It is important to note that the magnetic field strength is similar around the two poles of a magnet. But it is comparatively weaker in the center of the magnet. The pole strength varies inversely with the distance between the poles of the magnet. If the distance between two points is very large then the pole strength of the magnet will decrease.