Question

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. \(3.2 \times {10^{ - 18}}N\)
D. \(1.6 \times {10^{ - 6}}N\)

Answer 1

Hint: From the given energy of the proton we find the speed of the proton. When a proton is moving in the magnetic field then there is magnetic force acting on it. The magnitude of the magnetic force is proportional to the vector product of the velocity and the magnetic field.




Formula used:
\(\overrightarrow F = q\left( {\vec v \times \vec B} \right)\), here \(\vec F\)is the magnetic force vector, \(\vec v\)is the velocity of the charged particle and \(\vec B\)is the magnetic field in the region.
\(K = \frac{{m{v^2}}}{2}\), here K is the kinetic energy of the body having mass m moving with speed v.




Complete answer:
The energy of the proton is given as 200MeV
We convert the given energy into joules.
\(K = 200 \times {10^6} \times 1.6 \times {10^{ - 19}}J\)
\(K = 3.2 \times {10^{ - 11}}J\)
The mass of the proton is \(m = 1.67 \times {10^{ - 27}}Kg\)
Let the speed of the proton is v.
Then the kinetic energy of the proton is \(K = \frac{{m{v^2}}}{2}\)
\(\frac{{1.67 \times {{10}^{ - 27}}{v^2}}}{2} = 3.2 \times {10^{ - 11}}\)
\(v = \sqrt {\frac{{2 \times 3.2 \times {{10}^{ - 11}}}}{{1.67 \times {{10}^{ - 27}}}}} m/s\)
\(v = 1.96 \times {10^8}m/s\)
We assumed the Cartesian coordinate system equivalence to the geometric directions.
The magnetic field vector is \(\vec B = \left( {5T} \right)\hat j\)
The velocity vector is \(\vec v = \left( {1.96 \times {{10}^8}m/s} \right)\hat k\)
The magnetic force acting on the proton is,
\(\vec F = q\left( {\vec v \times \vec B} \right)\)
\(\vec F = 1.6 \times {10^{ - 19}}\left( {\left( {1.96 \times {{10}^8}\hat k} \right) \times \left( {5\hat j} \right)} \right)N\)
\(\vec F = \left( {1.57 \times {{10}^{ - 10}}N} \right)\hat- i\)
\(\vec F \approx \left( -{1.6 \times {{10}^{ - 10}}N} \right)\hat i\)
Hence, the magnitude of the magnetic force on the proton is approximately \(1.6 \times {10^{ - 10}}N\)
Therefore, the correct option is (B).







Note:We should be careful about the energy of the proton. We should consider the classical nature of the proton otherwise Lorentz's force law will not be valid for the subatomic particles like electrons and protons moving in the magnetic field.