A particle of mass m and charge q is in an el | vedclass

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(D) The total work done on the particle is the sum of the work done by the electric field and the magnetic field.Work done by the magnetic field on a

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$5$

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(D) The total work done on the particle is the sum of the work done by the electric field and the magnetic field.Work done by the magnetic field on a moving charge is always zero because the magnetic force $\vec{F}_m = q(\vec{v} 	imes \vec{B})$ is always perpendicular to the velocity $\vec{v}$.Thus,the total work done is equal to the work done by the electric field.The work done by the electric field is given by $W = \vec{F}_e \cdot \vec{d} = q \vec{E} \cdot \vec{d}$.The displacement vector is $\vec{d} = (1 - 0) \hat{i} + (1 - 0) \hat{j} = \hat{i} + \hat{j}$.Substituting the values: $W = q(2 \hat{i} + 3 \hat{j}) \cdot (\hat{i} + \hat{j})$.$W = q(2 	imes 1 + 3 	imes 1) = 5q$.Therefore,the magnitude of the total work done is $5q$.

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