Search any question & find its solution
Question:
Answered & Verified by Expert
The radius of the path of an electron moving at a speed of $3.2 \times 10^7 \mathrm{~ms}^{-1}$ in a magnetic field of $6 \times 10^{-4} \mathrm{~T}$ perpendicular to it is (mass of electron is $9 \times 10^{-31} \mathrm{~kg}$ and charge of electron is $\left.1.6 \times 10^{-19} \mathrm{C}\right)$
Options:
Solution:
2385 Upvotes
Verified Answer
The correct answer is:
$30 \mathrm{~cm}$
$\mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{9.1 \times 10^{-31} \times 3.2 \times 10^7}{1.6 \times 10^{-19} \times 6 \times 10^{-4}}=30 \mathrm{~cm}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.