
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 $1cm$ increased by $500gm$, the pole strength of each sphere is:
A) $1.4Am$
B) $9.8Am$
C) $0.7Am$
D) $70Am$
Answer
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Hint: 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 ${P_1}$ and ${P_2}$. Since their pole strengths are equal so
${P_1} = {P_2} = P$
Also the distance between the two spheres is $d = 1cm = 0.01m$

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
$F = \dfrac{{\mu {q_1}{q_2}}}{{4\pi {r^2}}}$
Where $F$ is the force
${q_1},{q_2}$are magnitude of charges, in this case it is ${P_1},{P_2}$
$\mu$ is the permeability
$r$ is the distance
Step III:
${F_1} = \dfrac{{\mu {P_1}{P_2}}}{{4\pi {d^2}}}$
Since the pole strengths are equal, so
${F_1} = \dfrac{{\mu {P^2}}}{{4\pi {d^2}}}$---(i)
Step IV: When the weight $500gm$ is added, then another force will act.
${F_2} = mg$
$m = 0.5kg;g = 10m/{s^2}$
${F_2} = 0.5 \times 10$---(ii)
Step V: Since the forces are balanced, so
${F_1} = {F_2}$
$\dfrac{{\mu {P^2}}}{{4\pi {d^2}}} = 0.5 \times 10$
Since the term $4\pi $, so substituting all the other values and solving
$\dfrac{{{{10}^{ - 7}}{P^2}}}{{{{(0.01)}^2}}} = 0.5 \times 10$
${P^2} = \dfrac{{0.5 \times 10 \times 0.0001}}{{{{10}^{ - 7}}}}$
${P^2} = 4998.49$
$P = 70.7Am$
Step VI: The pole strength of each sphere is $70.7Am.$
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.
Complete step by step solution:
Step I: Since the spheres are placed close to each other, so let their pole strengths be ${P_1}$ and ${P_2}$. Since their pole strengths are equal so
${P_1} = {P_2} = P$
Also the distance between the two spheres is $d = 1cm = 0.01m$

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
$F = \dfrac{{\mu {q_1}{q_2}}}{{4\pi {r^2}}}$
Where $F$ is the force
${q_1},{q_2}$are magnitude of charges, in this case it is ${P_1},{P_2}$
$\mu$ is the permeability
$r$ is the distance
Step III:
${F_1} = \dfrac{{\mu {P_1}{P_2}}}{{4\pi {d^2}}}$
Since the pole strengths are equal, so
${F_1} = \dfrac{{\mu {P^2}}}{{4\pi {d^2}}}$---(i)
Step IV: When the weight $500gm$ is added, then another force will act.
${F_2} = mg$
$m = 0.5kg;g = 10m/{s^2}$
${F_2} = 0.5 \times 10$---(ii)
Step V: Since the forces are balanced, so
${F_1} = {F_2}$
$\dfrac{{\mu {P^2}}}{{4\pi {d^2}}} = 0.5 \times 10$
Since the term $4\pi $, so substituting all the other values and solving
$\dfrac{{{{10}^{ - 7}}{P^2}}}{{{{(0.01)}^2}}} = 0.5 \times 10$
${P^2} = \dfrac{{0.5 \times 10 \times 0.0001}}{{{{10}^{ - 7}}}}$
${P^2} = 4998.49$
$P = 70.7Am$
Step VI: The pole strength of each sphere is $70.7Am.$
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.
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