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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(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 $5q$.

Similar Questions

If \(x_1\) be the size of the magnified image and \(x_2\) the size of the diminished image in Lens Displacement Method, then the size of the object is :

$A$ Carnot engine operating between temperatures $T_1$ and $T_2$ has an efficiency of $\frac{1}{6}$. When $T_2$ is lowered by $62\, K$,its efficiency increases to $\frac{1}{3}$. Then $T_1$ and $T_2$ are,respectively:

The direction of the angular velocity vector is along

The length of the perpendicular from the point $P(2, -1, 4)$ to the straight line $\frac{x+3}{10} = \frac{y-2}{-7} = \frac{z}{1}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module