Search any question & find its solution
Question:
Answered & Verified by Expert
A proton beam enters a magnetic field of \( 10^{-4} \mathrm{~Wb} m^{-2} \) normally. If the specific charge of the
proton is \( 10^{11} \mathrm{C} \mathrm{kg}^{-1} \) and its velocity is \( 10^{9} \mathrm{~ms}^{-1} \), then the radius of the circle described will
be
Options:
proton is \( 10^{11} \mathrm{C} \mathrm{kg}^{-1} \) and its velocity is \( 10^{9} \mathrm{~ms}^{-1} \), then the radius of the circle described will
be
Solution:
2977 Upvotes
Verified Answer
The correct answer is:
\( 100 \mathrm{~m} \)
Magnetic field \( B=10^{-4} \mathrm{~W} \mathrm{bm}^{-2} \), charge \( q=10^{11} \mathrm{Ck} g^{-1} \)
velocity \( v=10^{9} \mathrm{~ms}^{-1} \)
Now, radius of the circle described by proton is
\( r=\frac{m v}{q B}=\frac{v}{\left(\frac{q}{m}\right) \times B}=\frac{10^{9}}{10^{11} \times 10^{-4}}=10^{2} \)
Therefore, radius of circle is \( 100 \mathrm{~m} \).
velocity \( v=10^{9} \mathrm{~ms}^{-1} \)
Now, radius of the circle described by proton is
\( r=\frac{m v}{q B}=\frac{v}{\left(\frac{q}{m}\right) \times B}=\frac{10^{9}}{10^{11} \times 10^{-4}}=10^{2} \)
Therefore, radius of circle is \( 100 \mathrm{~m} \).
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.