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

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(B) Given: Velocity $v = (3 \hat{i} + 2 \hat{j}) \ m/s$,Magnetic field $B = (2 \hat{j} + 3 \hat{k}) \ T$,Specific charge $\frac{q}{m} = 0.96 	imes 10

Vedclass · Accepted Answer

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

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(B) Given: Velocity $v = (3 \hat{i} + 2 \hat{j}) \ m/s$,Magnetic field $B = (2 \hat{j} + 3 \hat{k}) \ T$,Specific charge $\frac{q}{m} = 0.96 	imes 10^8 \ C/kg$.The force acting on the proton is given by $F = q(v 	imes B)$.Calculate the cross product $v 	imes B$:$v 	imes B = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & 0 \ 0 & 2 & 3 \end{vmatrix} = \hat{i}(2 	imes 3 - 0 	imes 2) - \hat{j}(3 	imes 3 - 0 	imes 0) + \hat{k}(3 	imes 2 - 2 	imes 0) = 6 \hat{i} - 9 \hat{j} + 6 \hat{k}$.Using Newton's second law,$F = ma$,so $a = \frac{F}{m} = \frac{q}{m}(v 	imes B)$.Substituting the values:$a = (0.96 	imes 10^8) 	imes (6 \hat{i} - 9 \hat{j} + 6 \hat{k})$$a = 0.96 	imes 10^8 	imes 3 	imes (2 \hat{i} - 3 \hat{j} + 2 \hat{k})$$a = 2.88 	imes 10^8 	imes 100 	imes (2 \hat{i} - 3 \hat{j} + 2 \hat{k})$$a = 288 	imes 10^8(2 \hat{i} - 3 \hat{j} + 2 \hat{k}) \ m/s^2$.

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

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