An electron moving with a velocity vec{V_1} = | vedclass

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(A) The magnetic force on a charged particle is given by $\vec{F} = q(\vec{V} 	imes \vec{B})$.For an electron,$q = -e$.Given $\vec{V_1} = 2\hat{i}$ a

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(A) The magnetic force on a charged particle is given by $\vec{F} = q(\vec{V} 	imes \vec{B})$.For an electron,$q = -e$.Given $\vec{V_1} = 2\hat{i}$ and $\vec{F_1} = -2\hat{j}$,we have $-2\hat{j} = -e(2\hat{i} 	imes \vec{B}) \implies \hat{j} = e(\hat{i} 	imes \vec{B})$.Given $\vec{V_2} = 2\hat{j}$ and $\vec{F_2} = 2\hat{i}$,we have $2\hat{i} = -e(2\hat{j} 	imes \vec{B}) \implies \hat{i} = -e(\hat{j} 	imes \vec{B})$.Let $\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}$.From $\hat{i} 	imes (B_x\hat{i} + B_y\hat{j} + B_z\hat{k}) = \frac{1}{e}\hat{j}$,we get $B_z\hat{j} - B_y\hat{k} = \frac{1}{e}\hat{j}$,so $B_z = \frac{1}{e}$ and $B_y = 0$.From $\hat{j} 	imes (B_x\hat{i} + B_y\hat{j} + B_z\hat{k}) = -\frac{1}{e}\hat{i}$,we get $-B_z\hat{i} + B_x\hat{k} = -\frac{1}{e}\hat{i}$,so $B_z = \frac{1}{e}$ and $B_x = 0$.Thus,$\vec{B} = \frac{1}{e}\hat{k}$.Now,for $\vec{V_3} = 2\hat{k}$,the force is $\vec{F_3} = -e(2\hat{k} 	imes \frac{1}{e}\hat{k}) = -2(\hat{k} 	imes \hat{k}) = 0$.

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