A proton of velocity vec{v} = (3 hat{i} + 2 h | vedclass

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(B) Given: Velocity $\vec{v} = (3 \hat{i} + 2 \hat{j}) 	ext{ ms}^{-1}$, Magnetic field $\vec{B} = (2 \hat{j} + 3 \hat{k}) 	ext{ T}$.Specific charge

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$288 	imes 10^8(2 \hat{i} - 3 \hat{j} + 2 \hat{k})$

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(B) Given: Velocity $\vec{v} = (3 \hat{i} + 2 \hat{j}) 	ext{ ms}^{-1}$, Magnetic field $\vec{B} = (2 \hat{j} + 3 \hat{k}) 	ext{ T}$.Specific charge $\frac{q}{m} = 0.96 	imes 10^8 	ext{ C kg}^{-1}$.The magnetic force on a moving charge is $\vec{F} = q(\vec{v} 	imes \vec{B})$.According to Newton's second law, $\vec{F} = m\vec{a}$, so $\vec{a} = \frac{q}{m}(\vec{v} 	imes \vec{B})$.First, calculate the cross product $\vec{v} 	imes \vec{B}$:$\vec{v} 	imes \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & 0 \ 0 & 2 & 3 \end{vmatrix} = \hat{i}(2 	imes 3 - 0 	imes 2) - \hat{j}(3 	imes 3 - 0 	imes 0) + \hat{k}(3 	imes 2 - 2 	imes 0) = 6 \hat{i} - 9 \hat{j} + 6 \hat{k}$.Now, calculate acceleration $\vec{a} = \frac{q}{m}(6 \hat{i} - 9 \hat{j} + 6 \hat{k})$:$\vec{a} = 0.96 	imes 10^8 	imes (6 \hat{i} - 9 \hat{j} + 6 \hat{k})$$\vec{a} = 0.96 	imes 10^8 	imes 3 	imes (2 \hat{i} - 3 \hat{j} + 2 \hat{k})$$\vec{a} = 2.88 	imes 10^8 	imes (2 \hat{i} - 3 \hat{j} + 2 \hat{k}) = 288 	imes 10^6 	imes (2 \hat{i} - 3 \hat{j} + 2 \hat{k}) = 288 	imes 10^8 	imes (2 \hat{i} - 3 \hat{j} + 2 \hat{k}) 	ext{ ms}^{-2}$ (Note: The numerical coefficient matches option $B$).

$A$ proton of velocity $\vec{v} = (3 \hat{i} + 2 \hat{j}) \text{ ms}^{-1}$ enters a magnetic field of induction $\vec{B} = (2 \hat{j} + 3 \hat{k}) \text{ T}$. The acceleration produced in the proton in $\text{ms}^{-2}$ is (Specific charge of proton $= 0.96 \times 10^8 \text{ C kg}^{-1}$)

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