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Question: The force between two magnetic poles reduces to \(aN\), if the distance between them is increased to...

The force between two magnetic poles reduces to aNaN, if the distance between them is increased to nn times and it increases to bNbN if the distance between them is 1nth\dfrac{1}{{{n^{th}}}} of the original value. Then a:ba:b is
A. 1:n21:{n^2}
B. n2:1{n^2}:1
C. n4:1{n^4}:1
D. 1:n41:{n^4}

Explanation

Solution

We know that the force between two magnetic poles is classically given by the equation F=μ0m1m24πr2F = \dfrac{{{\mu _0}{m_1}{m_2}}}{{4\pi {r^2}}} . Use this equation for when the distance increases by nn times and when the distance reduces by 1nth\dfrac{1}{{{n^{th}}}} times. Then compare these equations to reach the solution.

Complete step by step answer:
If both poles are small enough to be described as single points, then magnetic charges can be regarded as point charges. The force between two magnetic poles is classically given by the equation
F=μ0m1m24πr2F = \dfrac{{{\mu _0}{m_1}{m_2}}}{{4\pi {r^2}}}
Here, F=F = The magnetic force between the two magnetic poles
μ0={\mu _0} = Magnetic susceptibility of free space
m1={m_1} = The magnetic moment of one magnetic pole
m2={m_2} = The magnetic moment of the second magnetic pole
r=r = The distance between the two magnetic poles
This equation is an example of an inverse square law
So, let the initial distance between the two magnetic poles be dd and the initial force be FF
So, for the initial position of the magnetic poles
F=μ0m1m24πd2F = \dfrac{{{\mu _0}{m_1}{m_2}}}{{4\pi {d^2}}}
When the distance between the two magnetic poles is increased by nn times then the magnetic force between the two magnetic poles becomes aNaN
F1=aN=μ0m1m24π(nd)2{F_1} = aN = \dfrac{{{\mu _0}{m_1}{m_2}}}{{4\pi {{\left( {nd} \right)}^2}}} (Equation 1)
When the distance between the two magnetic poles is reduced by 1n\dfrac{1}{n} times then the magnetic force between the two magnetic poles becomes bNbN
F2=bN=μ0m1m24π(dn)2{F_2} = bN = \dfrac{{{\mu _0}{m_1}{m_2}}}{{4\pi {{\left( {\dfrac{d}{n}} \right)}^2}}} (Equation 2)
Dividing equation 2 by equation 1, we get
aNbN=μ0m1m24π(nd)2μ0m1m24π(dn)2\dfrac{{aN}}{{bN}} = \dfrac{{\dfrac{{{\mu _0}{m_1}{m_2}}}{{4\pi {{\left( {nd} \right)}^2}}}}}{{\dfrac{{{\mu _0}{m_1}{m_2}}}{{4\pi {{\left( {\dfrac{d}{n}} \right)}^2}}}}}
aNbN=1n4\dfrac{{aN}}{{bN}} = \dfrac{1}{{{n^4}}}
a:b=1:n4\therefore a:b = 1:{n^4}

So, the correct answer is “Option D”.

Note:
Due to their motion, magnetic force, attraction, or repulsion that occurs between electrically charged particles. It is the underlying force responsible for phenomena such as the operation of electric motors and the iron attraction of magnets. Among stationary electric charges, electric forces exist; among moving electric charges, both electric and magnetic forces exist.