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

The net force on the charged particle is given by:

𝐅net = q𝐄 + q(𝐯 × 𝐁)

where 𝐄 is the electric field, 𝐁 is the magnetic field, and 𝐯 is the velocity of the particle.

The work done (W) in moving the particle from the origin to point P(1,1) along a straight path is given by the dot product of the net force and the displacement vector:

W = 𝐅net ⋅ 𝐒

Since the displacement is along a straight path from (0,0) to (1,1), the displacement vector 𝐒 = î + ĵ.

The work done by the magnetic force is zero because the magnetic force is always perpendicular to the velocity of the charged particle (and hence perpendicular to the displacement vector). Therefore, only the electric force contributes to the work done.

The electric force is:

𝐅electric = q𝐄 = q(2î + 3ĵ)

Therefore, the work done is:

W = 𝐅electric ⋅ 𝐒 = q(2î + 3ĵ) ⋅ (î + ĵ) = q(2(1) + 3(1)) = 5q

Therefore, the magnitude of the total work done is 5q.