JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A particle having the same charge as of electron moves in a circular path of radius \(0.5
\,cm\) under the influence of a magnetic field of \(0.5\,T.\) If an electric field of \(100\,V/m\) makes it to move in a straight path, then the mass of the particle is (given charge of electron \(= 1.6 \times 10^{-19}\, C\) )
- A \(9.1 \times 10^{-31}\, kg\)
- B \(1.6 \times 10^{-27}\, kg\)
- C \(1.6 \times 10^{-19}\, kg\)
- D \(2.0 \times 10^{-24}\, kg\)
Answer & Solution
Correct Answer
(D) \(2.0 \times 10^{-24}\, kg\)
Step-by-step Solution
Detailed explanation
\(\mathrm{eE}=\mathrm{evB}\) \(\Rightarrow E=\left(\frac{e B r}{m}\right) B\) \(\Rightarrow m=\frac{e B^{2} r}{E}\) \(\Rightarrow \mathrm{m}=\frac{\left(1.6 \times 10^{-10}\right)(0.5)^{2}\left(0.5 \times 10^{-2}\right)}{100}=2 \times 10^{-24}\, \mathrm{kg}\)
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