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  • A magnetic needle lying parallel to a magnetic field requires a units of work to turn it through <img alt="30^{\circ}" src="https://entrancecorner.oncodecogs.com/gif.

A magnetic needle lying parallel to a magnetic field requires a units of work to turn it through 30^{\circ}. The torque needed to maintain the needle in this position will be

Option: 1

\frac{a}{2-\sqrt{3}}


Option: 2

\frac{a}{\sqrt{3-2}}


Option: 3

\frac{a}{2}


Option: 4

-\frac{a}{2}


Answers (1)

best_answer

\begin{aligned} \text { work }=-M B\left(\cos \theta_2 \cos \theta_2\right) \\ \end{aligned}

\begin{aligned} & \Rightarrow \quad a=-M B\left(\cos 30^{\circ}-\cos 0^{\circ}\right) \\\end{aligned}

\begin{aligned} & \Rightarrow \quad a=-M B \times\left(\frac{\sqrt{3}}{2}\right)=-\frac{\sqrt{3} M B}{2}+M B \\\\\end{aligned}                                                                                    ------(1)

                                                      \begin{aligned} & =\frac{(2-\sqrt{3}) B M}{2} \\\end{aligned}

so  \begin{aligned} a=M B \sin 30^{\circ} \\\end{aligned}

         \begin{aligned} =\frac{-2 a}{(2-\sqrt{3})} \times \frac{1}{2} \\\end{aligned}    (From equation (1))

         \begin{aligned} =\frac{a}{(-\sqrt{3}+2)}=\frac{a}{\sqrt{3}-2} \\\end{aligned}, (Left side of                                                                          equilibrium)

 

Posted by

vishal kumar

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