A proton of velocity 3i^+2j^ms−1 enters a field of magnetic induction 2j^+3k^ tesla. The acceleration produced in the proton is (charge to mass ratio of proton = 0.96 × 108 C kg-1)

F→=qV→×B→a=Fm=qmV×B=ijk320023qm=0.96×1086i−9j+6k=2.88×1082i−3j+2k

A proton of velocity 3i^+2j^ms−1 enters a field of magnetic induction 2j^+3k^ tesla. The acceleration produced in the proton is (charge to mass ratio of proton = 0.96 × 108 C kg-1)

F→=qV→×B→a=Fm=qmV×B=ijk320023qm=0.96×1086i−9j+6k=2.88×1082i−3j+2k

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A proton of velocity $(3 \hat{i} + 2 \hat{j}) m s^{- 1}$ enters a field of magnetic induction $(2 \hat{j} + 3 \hat{k})$ tesla. The acceleration produced in the proton is (charge to mass ratio of proton = $0.96 \times 10^{8} C k g^{- 1}$ )

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$2.8 \times 10^{8} (2 \hat{i} - 3 \hat{j})$

$2.88 \times 10^{8} (2 \hat{i} - 3 \hat{j} + 2 \hat{k})$

$2.8 \times 10^{8} (2 \hat{i} + 3 \hat{k})$

$2.88 \times 10^{8} (\hat{i} - 3 \hat{j} + 2 \hat{k})$

answer is B.

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Detailed Solution

$\begin{array}{l} \vec{F} = q (\vec{V} \times \vec{B}) \\ a = \frac{F}{m} = \frac{q}{m} (V \times B) = |\begin{matrix} i & j & k \\ 3 & 2 & 0 \\ 0 & 2 & 3 \end{matrix}| \frac{q}{m} \end{array}$

$\begin{array}{l} = 0.96 \times 10^{8} [6 i - 9 j + 6 k] \\ = 2.88 \times 10^{8} [2 i - 3 j + 2 k] \end{array}$

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