Question

A particle of mass m and charge q is in an electric and magnetic field given by $\vec E = 2\hat i + 3\hat j;\vec B = 4\hat j + 6\hat k.$ 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?
A. (0.35) q
B. (0.15) q
C. (2.5) q
D. 5q

Answer 1

- Hint: In this question we are being asked about the magnitude of the total work done, as we can see that work done(w)= f.s where f is considered force and s is stated as displacement in the direction of force. Now by placing Lorentz force$F = q\vec E + q(\vec v \times \vec B)$, we can find our required answer.

Complete step-by-step solution -

We know that,
Work done = force $\times$ displacement in the direction of force (w=f.s)
Now we need to find the force on charge of mass m,
Force applied on any charge that is kept on electric or magnetic field
$F = q\vec E + q(\vec v \times \vec B)$ (Lorentz force)
It is stated in the question that the particle is being shifted from origin to point p as we can conclude that the particle does not have its own velocity so,
P ($x\hat i + y\hat j$), as x and y = 1
So, coordinates of P point would be,
P ($\hat i + \hat j$)
The particle is being shifted from origin to point p so,
$q(\vec v \times \vec B) = 0$,
Now force f = $q\vec E$ ($\vec E = 2\hat i + 3\hat j$)
We need to multiply force with displacement and
The direction of force is along $\hat i + \hat j$ and displacement is also along $\hat i + \hat j$ , force and displacement would have ${0^0}$ angle
So, work done W= f.s
W= q ($2\hat i + 3\hat j$) ($\hat i + \hat j$)
Q (2+3) = 5q
Hence, the magnitude of the total work done is 5q.
Option d is the correct answer.

Note- In this question, we saw that the particle was at rest at origin and we did some external work in shifting it from there to the point P. This work was done against the electric and magnetic fields. Therefore, we calculated the respective forces and hence the total work done.