OABC is a current-carrying square loop. An el | vedclass

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(D) The magnetic field $\vec{B}$ at the center of the square loop is directed perpendicular to the plane of the loop,pointing outwards (using the righ

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$\frac{\hat{i} + \hat{j}}{\sqrt{2}}$

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(D) The magnetic field $\vec{B}$ at the center of the square loop is directed perpendicular to the plane of the loop,pointing outwards (using the right-hand thumb rule),so $\vec{B} = B\hat{k}$.The electron is projected along the diagonal $AC$ towards $C$. The direction of velocity $\vec{v}$ is along the vector from the center to $C$,which is in the direction of $(\hat{i} - \hat{j})$.The Lorentz force on the electron is given by $\vec{F} = -e(\vec{v} 	imes \vec{B})$.Substituting the values: $\vec{F} = -e [v(\frac{\hat{i} - \hat{j}}{\sqrt{2}}) 	imes B\hat{k}] = -evB [\frac{1}{\sqrt{2}}(\hat{i} 	imes \hat{k} - \hat{j} 	imes \hat{k})]$.Since $\hat{i} 	imes \hat{k} = -\hat{j}$ and $\hat{j} 	imes \hat{k} = \hat{i}$,we get $\vec{F} = -evB [\frac{1}{\sqrt{2}}(-\hat{j} - \hat{i})] = \frac{evB}{\sqrt{2}}(\hat{i} + \hat{j})$.The acceleration $\vec{a} = \frac{\vec{F}}{m_e}$ is in the same direction as the force $\vec{F}$.Thus,the unit vector in the direction of initial acceleration is $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$.

$OABC$ is a current-carrying square loop. An electron is projected from the center of the loop along its diagonal $AC$ as shown. The unit vector in the direction of initial acceleration will be

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