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/2sqrt(2W"tan"alpha/2)`
B. `2l"sin"alpha/2sqrt(2W"tan"alpha/2)`
C. `3l"sin"alpha/2sqrt(2W"tan"alpha/2)`
D. `4l"sin"alpha/2sqrt(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 repulsive 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.,
`Wl/2"sin"alpha/2=Fl"cos"alpha/2`
Hence `F=m^(2)/a^(2)," where "a=2l"sin"alpha/2`
Thus `W/2"sin"alpha/2=(m^(2)"cos"alpha/2)/(4l^(2)"sin"alpha/2)`
`m=l"sin"alpha/2sqrt(2W"tan"alpha/2)`