MHT CET · Physics · Magnetic Effects of Current
Electron of mass ' \(m\) ' and charge ' \(q\) ' is travelling with speed ' \(v\) ' along a circular path of radius ' \(R\) ' at right angles to a uniform magnetic field of intensity ' \(B\) '. If the speed of the electron is halved and the magnetic field is doubled, the resulting path would have radius
- A \(\frac{\mathrm{R}}{2}\)
- B \(\frac{\mathrm{R}}{4}\)
- C 2 R
- D 4 R
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{R}}{4}\)
Step-by-step Solution
Detailed explanation
From cyclotron motion and uniform circular motion,
i.e., \(\mathrm{q} v \mathrm{~B} \sin 90^{\circ}=\frac{\mathrm{mv}^2}{\mathrm{R}}\)
\(\therefore \quad \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}\)...(i)
Given \(\mathrm{v}^{\prime}=\frac{\mathrm{v}}{2}\) and \(\mathrm{B}^{\prime}=2 \mathrm{~B}\)
\(\begin{aligned}
& \therefore \quad R^{\prime}=\frac{\mathrm{mv}^{\prime}}{\mathrm{qB}^{\prime}}=\frac{\mathrm{m} \frac{\mathrm{v}}{2}}{\mathrm{q} 2 \mathrm{~B}}=\frac{1}{4} \frac{\mathrm{mv}}{\mathrm{qB}} \\
& \quad \Rightarrow R^{\prime}=\frac{1}{4} R
\end{aligned}\)
...[From(i)]
i.e., \(\mathrm{q} v \mathrm{~B} \sin 90^{\circ}=\frac{\mathrm{mv}^2}{\mathrm{R}}\)
\(\therefore \quad \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}\)...(i)
Given \(\mathrm{v}^{\prime}=\frac{\mathrm{v}}{2}\) and \(\mathrm{B}^{\prime}=2 \mathrm{~B}\)
\(\begin{aligned}
& \therefore \quad R^{\prime}=\frac{\mathrm{mv}^{\prime}}{\mathrm{qB}^{\prime}}=\frac{\mathrm{m} \frac{\mathrm{v}}{2}}{\mathrm{q} 2 \mathrm{~B}}=\frac{1}{4} \frac{\mathrm{mv}}{\mathrm{qB}} \\
& \quad \Rightarrow R^{\prime}=\frac{1}{4} R
\end{aligned}\)
...[From(i)]
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