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A proton is projected with a speed of $4 × 10^6\, m s^{-1}$ horizontally from east to west. A uniform magnetic field of strength $3.0 × 10^{-3} T$ exists in the vertically upward direction. The acceleration of the proton is (Given: mass of proton $1.67 × 10^{-27} kg$; charge on proton $1.6 × 10^{-19} C$) |
zero $11.5 × 10^{11}\, m s^{-2}$ $18.3 × 10^{10}\, m s^{-2}$ $32.1 × 10^{11}\, m s^{-2}$ |
$11.5 × 10^{11}\, m s^{-2}$ |
The correct answer is Option (2) → $11.5 × 10^{11}\, m s^{-2}$ |
