A proton is projected horizontally from east to west in a uniform magnetic field of strength \( 3.0\times10^{-3}\,\mathrm{T} \), with a speed of \( 2\times10^{6}\,\mathrm{m\,s^{-1}} \). The magnetic field exists in vertically upward direction, then the force on proton due to the magnetic field will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 9.6\times10^{-16}\,\mathrm{N} \) towards north
The magnetic force on a moving charge is \( \vec{F} = q(\vec{v}\times\vec{B}) \). Here \( \vec{v} \) is perpendicular to \( \vec{B} \), so \( \sin\theta = 1 \).
Magnitude
\( F = qvB = 1.6\times10^{-19}\times2\times10^{6}\times3.0\times10^{-3} \)
\( F = 9.6\times10^{-16}\,\mathrm{N} \)
Direction (east = \( +\hat{x} \), north = \( +\hat{y} \), up = \( +\hat{z} \))
\( \vec{v} = -v\hat{x} \) (east to west), \( \vec{B} = +B\hat{z} \)
\( \vec{v}\times\vec{B} = -vB(\hat{x}\times\hat{z}) = -vB(-\hat{y}) = +vB\,\hat{y} \)
The proton is positive, so the force is along \( +\hat{y} \), that is, towards the north.