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Question: Two short magnets have equal pole strengths but one is twice as long as the other. The shorter magne...

Two short magnets have equal pole strengths but one is twice as long as the other. The shorter magnet is placed 20cm20\,cm in tanA\tan \,A position from the compass needle. The longer magnet must be placed on the other side of the magnetometer for no deflection at a distance equal to
(A) 20cm20\,cm
(B) 20×(2)1/3cm20 \times {\left( 2 \right)^{1/3}}\,cm
(C) 20×(2)2/3cm20 \times {\left( 2 \right)^{2/3}}\,cm
(D) 20×(2)3/3cm20 \times {\left( 2 \right)^{3/3}}\,cm

Explanation

Solution

Use the condition for the no deflection between the magnets, frame the relation between the length and the distance of the magnet from the compass needle. Substitute the known values to find the value of the distance of the second magnet from the compass needle.

Useful formula:
The formula of the no deflection in the position of the tanA\tan \,A from the compass needle is
2M1μ04πd13=2M2μ04πd23\dfrac{{2{M_1}{\mu _0}}}{{4\pi {d_1}^3}} = \dfrac{{2{M_2}{\mu _0}}}{{4\pi {d_2}^3}}
Where M1{M_1} is the length of the first magnet, M2{M_2} is the length of the first magnet, d1{d_1} is the distance of the first magnet from the compass needle and d2{d_2} is the distance of the second magnet from the compass needle.

Complete step by step solution:
It is given that the
The shorter magnet is placed at a distance from tanA\tan \,A , d1=20cm{d_1} = 20\,cm
The longer magnet is twice the length of the shorter magnet.
Using the formula of the no deflection,
2M1μ04πd13=2M2μ04πd23\dfrac{{2{M_1}{\mu _0}}}{{4\pi {d_1}^3}} = \dfrac{{2{M_2}{\mu _0}}}{{4\pi {d_2}^3}}
By cancelling the similar terms on both sides,
M1d13=M2d23\dfrac{{{M_1}}}{{{d_1}^3}} = \dfrac{{{M_2}}}{{{d_2}^3}}
By rearranging the terms in the above step, we get
M1M2=d13d23\dfrac{{{M_1}}}{{{M_2}}} = \dfrac{{{d_1}^3}}{{{d_2}^3}}
Substituting the known values in the above equation,
12=203d23\dfrac{1}{2} = \dfrac{{{{20}^3}}}{{{d_2}^3}}
By simplifying the above values,
d23=4000{d_{{2^{}}}}^3 = 4000
Hence the value of the distance of the second magnet from the compass needle is obtained as follows.
d23=20×(2)1/3cm{d_2}^3 = 20 \times {\left( 2 \right)^{1/3}}\,cm

Thus the option (B) is correct.

Note: In the question, it is given that the Two short magnets have equal pole strengths but one is twice as long as the other. Hence in the above calculation, the value of the length of the first magnet is taken as 11 and that of the second magnet is 2(1)=22\left( 1 \right) = 2.