KCET · Physics · Electromagnetic Induction
A charge \( q \) is accelerated through a potential difference \( V \). It is then passed normally through
a uniform magnetic field, where it moves in a circle of radius \( \mathrm{r} \). The potential difference
required to move it in a circle of radius \( 2 r \) is
- A \( 12 \mathrm{~V} \)
- B 4V
- C \( 1 \mathrm{~V} \)
- D \( 3 \mathrm{~V} \)
Answer & Solution
Correct Answer
(B) 4V
Step-by-step Solution
Detailed explanation
Given, charge \(\mathrm{q}\) is accelerated through a potential difference \(\mathrm{V}\). Therefore,
\(\frac{1}{2} m v^{2}=q V\)
\(\Rightarrow m v^{2}=2 q V\)
\(\Rightarrow(m v)^{2}=2 q m V\)
\(\Rightarrow m v=\sqrt{2 m q V}\)
The radius of circular path followed by the charge is
\(r=\frac{m v}{q B}=\frac{\sqrt{2 m q V}}{q B}\)
\(\Rightarrow r \propto \sqrt{V}\)
Thus, the potential difference required to move the charge in circle of radius \(2 r\) is \(4 \mathrm{~V}\)
\(\frac{1}{2} m v^{2}=q V\)
\(\Rightarrow m v^{2}=2 q V\)
\(\Rightarrow(m v)^{2}=2 q m V\)
\(\Rightarrow m v=\sqrt{2 m q V}\)
The radius of circular path followed by the charge is
\(r=\frac{m v}{q B}=\frac{\sqrt{2 m q V}}{q B}\)
\(\Rightarrow r \propto \sqrt{V}\)
Thus, the potential difference required to move the charge in circle of radius \(2 r\) is \(4 \mathrm{~V}\)
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