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A particle of mass m and charge q is in an electric and magnetic field given by E = 2i + 3j; B = 4j + 6k. The charged particle is shifted from he origin to the point P(x = 1; y = 1) along a straight path. The magnitude of the total work done is:

1. (2.5) q
2. (0.35) q
3. (0.15) q
4. 5q

Correct answer: 5q

Solution

The magnetic force is always perpendicular to velocity and does no work, so only the electric force contributes: W = qE.d = q(2i+3j).(i+j) = q(2+3) = 5q.

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