Two long equally magnetized needles are freely suspended by their like poles form a hooks shown in figure. The length of each needle is `l cm` and the weight is `W.` in equillibrium the needles make an angle `alpha` with each other. The magnetic pole strength is concentrated at the ends of needles. The magnetic pole strength of the needles is

Options

A. `l sin((alpha)/(2))sqrt(2W tan((alpha)/(2)))`
B. `2lsin((alpha)/(2))sqrt(2W tan((alpha)/(2)))`
C. `3lsin((alpha)/(2))sqrt(2W tan((alpha)/(2)))`
D. `4lsin((alpha)/(2))sqrt(2W tan((alpha)/(2)))`

Answer

A

Explanation

Let us consider one of the needles. It will be acted by the weight `W`, which acts through the centre of gravity `B` and the replusive force between the poles `F=(m^(2))/(a^(2))`acted through the point `A`. For the needle to be in equilibrium the sum of the moments of forces acting on the needle should be equal to zero, i.e.,
`W(l)/(2)sin((alpha)/(2))=Flcos((alpha)/(2))`
Hence `F=(m^(2))/(a^(2))`, where `a=2lsin((alpha)/(2))`
Thus `(W)/(2)sin((alpha)/(2))=m^(2)cos((alpha)/(2))/(4l^(2)sin^(2)((alpha)/(2)))`
`m=l sin((alpha)/(2))sqrt(2W tan((alpha)/(2)))`