Question

A proton moving with a velocity \[2.5 \times {10^7}m/s\]enters a magnetic field of intensity 2.5T making an angle \[30^\circ \]with the magnetic field. The force on the proton is
A. \[3 \times {10^{ - 12}}N\]
B. \[5 \times {10^{ - 12}}N\]
C. \[6 \times {10^{ - 12}}N\]
D. \[9 \times {10^{ - 12}}N\]

Answer 1

Hint: When the proton is moving in a magnetic field making an angle with the magnetic field vector, then the vertical component will cause the magnetic force on the proton, i.e. we need to find the vector product of the velocity and the magnetic field.



\[\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)\], 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.





Complete answer:
The charge on the proton is positive and the magnitude of the charge is \[1.6 \times {10^{ - 19}}C\]
\[q = + 1.6 \times {10^{ - 19}}C\]
The speed of the motion is given as \[2.5 \times {10^7}m/s\]
The magnetic field strength in the region is given uniform and equal to 2.5T
\[B = 2.5T\]
It is given that the angle between the velocity vector and the magnetic field is \[30^\circ \]
\[\theta = 30^\circ \]
Using the Lorentz’s force law, the magnetic force on the proton due to interaction with magnetic field in the region is,
\[\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)\]
Then the magnitude of the magnetic force on the proton is,
\[\left| F \right| = \left| {qvB\sin \theta } \right|\]
Putting the values in the expression for the magnitude of the magnetic field on the proton, we get
\[\left| F \right| = 1.6 \times {10^{ - 19}} \times 2.5 \times {10^7} \times 2.5 \times \left( {\sin 30^\circ } \right)N\]
\[\left| F \right| = 5 \times {10^{ - 12}}N\]
So, when the proton enters into the region of the magnetic field then it experiences magnetic force of magnitude \[5 \times {10^{ - 12}}N\]
Therefore, the correct option is (B).





Note:As there is no electric field in the region, so in Lorentz’s law of force formula, we take the electric field as zero. The magnitude of the magnetic force doesn’t depend on the nature of the charge. If instead of a proton, it was an electron the magnitude of the magnetic force would be the same.