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(C) The time period of the circular motion of a charged particle in a magnetic field is $T = \frac{2\pi m}{qB}$.Given time $t = \frac{\pi m}{qB} = \fr

Vedclass · Accepted Answer

$-6\hat{k} + 8\hat{j}$ units

Vedclass · Answer

(C) The time period of the circular motion of a charged particle in a magnetic field is $T = \frac{2\pi m}{qB}$.Given time $t = \frac{\pi m}{qB} = \frac{T}{2}$.In a time interval of $\frac{T}{2}$,the velocity component perpendicular to the magnetic field $\vec{B}$ is reversed.The initial velocity is $\vec{v}_0 = 2\hat{i} + 3\hat{j} + 4\hat{k}$.The component parallel to $\vec{B} = 2\hat{i}$ is $v_{\parallel} = 2\hat{i}$.The perpendicular component is $\vec{v}_{\perp} = 3\hat{j} + 4\hat{k}$.After time $t = \frac{T}{2}$,the velocity becomes $\vec{v} = v_{\parallel} - \vec{v}_{\perp} = 2\hat{i} - 3\hat{j} - 4\hat{k}$.For the particle to move in a straight line with constant speed,the net Lorentz force must be zero: $\vec{F} = q(\vec{E} + \vec{v} 	imes \vec{B}) = 0$.Thus,$\vec{E} = -(\vec{v} 	imes \vec{B}) = -((2\hat{i} - 3\hat{j} - 4\hat{k}) 	imes 2\hat{i})$.Using the cross product rules $\hat{i} 	imes \hat{i} = 0$,$\hat{j} 	imes \hat{i} = -\hat{k}$,and $\hat{k} 	imes \hat{i} = \hat{j}$:$\vec{E} = -[0 - 3(-\hat{k}) - 4(\hat{j})] = -[3\hat{k} - 4\hat{j}] = -3\hat{k} + 4\hat{j}$.Wait,re-calculating: $\vec{E} = -[2\hat{i} 	imes 2\hat{i} - 3\hat{j} 	imes 2\hat{i} - 4\hat{k} 	imes 2\hat{i}] = -[0 - 6(-\hat{k}) - 8(\hat{j})] = -[6\hat{k} - 8\hat{j}] = -6\hat{k} + 8\hat{j}$ units.

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