Practicing Success
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A proton is projected horizontally from east to west in a uniform magnetic field of strength $3.0 \times 10^{-3} T$, with a speed of $2 \times 10^6 ms^{-1}$. The magnetic field exists in vertically upward direction, then the force on proton due to the magnetic field will be: |
$9.6 \times 10^{-16} N$ towards north $9.6 \times 10^{-16} N$ towards south $4.8 \times 10^{-16} N$ vertically upward $2.4 \times 10^{-28} N$ towards south |
$9.6 \times 10^{-16} N$ towards north |
The correct answer is Option (1) → $9.6 \times 10^{-16} N$ towards north |
