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  • <img alt="\begin{array}{l}{\text { A thin non-conducting ring of mass } m \text { carrying a charge } q} \\ {\text { can freely rotate about its axis. At the initial moment, the ring }}\\ {\text {

\begin{array}{l}{\text { A thin non-conducting ring of mass } m \text { carrying a charge } q} \\ {\text { can freely rotate about its axis. At the initial moment, the ring }}\\ {\text { was at rest and no magnetic field was present. Then a uniform }} \\ {\text { magnetic field was switched on, which was perpendicular to }} \\ {\text { the plane of the ring and increased with time according to a }}\end{array}

\text { certain } \operatorname{law}: \frac{d B}{d t}=k

\text { Find the angular velocity } \omega \text { of the ring as a function of } k

Option: 1

\omega=\frac{k q}{2 m} t


Option: 2

\omega=\frac{k m}{2 q} t


Option: 3

\omega=\frac{k q}{2 mt}


Option: 4

\omega=\frac{ q}{2k m} t


Answers (1)

best_answer

 

 

 

 

E=\frac{1}{2} R \frac{d B}{d t}

\begin{array}{l}{\text { Electric force on charge } d q \text { is given by }} \\ {\qquad d F=E d q=\frac{1}{2} R\left(\frac{d B}{d t}\right) d q} \\ {\Rightarrow \quad d \tau=R d F}\end{array}

\begin{array}{l}{\Rightarrow \quad d \tau=\frac{1}{2} R^{2}\left(\frac{d B}{d t}\right) d q} \\ {\Rightarrow \quad \tau=\frac{1}{2} R^{2} k q} \\ {\text { Now, } \tau=\frac{1}{2} R^{2} k q} \\ \\ {\Rightarrow \quad I \alpha=\frac{1}{2} R^{2} k q \quad \Rightarrow \quad m R^{2} \alpha=\frac{1}{2} R^{2} k q} \\ \\ {\quad \alpha=\frac{k q}{2 m} \quad \Rightarrow \quad \omega=\frac{k q}{2 m} t}\end{array}

Posted by

Kuldeep Maurya

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