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4.Moving Charges and Magnetism
medium
A proton of energy $200\, MeV$ enters the magnetic field of $5\, T$. If direction of field is from south to north and motion is upward, the force acting on it will be
A
Zero
B
$1.6 \times {10^{ - 10}}\,N$
C
$3.2 \times {10^{ - 8}}\,N$
D
$1.6 \times {10^{ - 6}}\,N$
Solution
(b) $F = qvB$ also Kinetic energy $K = \frac{1}{2}m{v^2}$$ \Rightarrow v = \sqrt {\frac{{2K}}{m}} $
$F = q\sqrt {\frac{{2K}}{m}} B$
$ = 1.6 \times {10^{ – 19}}\sqrt {\frac{{2 \times 200 \times {{10}^6} \times 1.6 \times {{10}^{ – 19}}}}{{1.67 \times {{10}^{ – 27}}}}} \times 5$
$ = 1.6 \times {10^{ – 10}}\,N$
Standard 12
Physics
