Question

A circular coil has a moment of inertia$0.8kg{m}^2$ around any diameter and is carrying the current to produce a magnetic moment of $20A{m}^2$. The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of $4T$ is applied along the vertical, it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by$60°$ will be:

• A. $10\pi rad{s}^{-1}$
• B. $20rad{s}^{-1}$
• C. $20\pi rad{s}^{-1}$
• D. $10rad{s}^{-1}$

Answer 1

The correct option is D

$10rad{s}^{-1}$

Step 1 : Given data

Moment of inertia$=$ $0.8kg{m}^2$

Magnetic moment$=$$20A{m}^2$

Magnetic field $=$$4T$

Angle $\theta ={60}^0$

Angular speed $=\omega$

Step 2: Formula used

Rotational kinetic energy of circular coil $KE=\frac{1}{2}I{\omega }^2$

Potential energy of coil , when considered equivalent to a bar magnet is

$U=-MB\cos \theta$

Where, $M=$Magnetic momentum

$I=$Moment of inertia

$B=$Magnetic field

Initial potential energy $=U_i=-MB\cos 60°$

Initial kinetic energy $=K_i$

Final potential energy $=U_f=-MB$

Final kinetic energy $=K_f$

Step 3:To find the angular speed $\left(\omega \right)$

According to law of energy conservation

$\frac{1}{2}I{\omega }^2=U_i-U_f$

$\Rightarrow$ $$\begin{gathered} \frac{1}{2}I{\omega }^2=-MB\cos \theta -\left(-MB\right) \\ =-MB\cos {60}^0-\left(-MB\right) \end{gathered}$$

$\Rightarrow$ $\frac{MB}{2}=\frac{1}{2}I{\omega }^2$

$\Rightarrow$$\frac{20\times 4}{2}=\frac{1}{2}\left(0.8\right){\omega }^2$

$\Rightarrow$ $\begin{array}{rcl}\omega & =& \sqrt{100}\\ & =& 10rad{s}^{-1}\end{array}$

Therefore, the angular speed is $\mathbf{10}\mathbf{}\mathit{r}\mathit{a}\mathit{d}\mathbf{}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$

Hence, The correct answer is Option D