A particle of mass $m$ and charge $q$ has an initial velocity $\vec{v}=v_{0} \hat{j}$. If an electric field $\vec{E}=E_{0} \hat{i}$ and magnetic field $\vec{B}=B_{0} \hat{i}$ act on the particle, its speed will double after a time:

(C) $\frac{\sqrt{3} m v_{0}}{q E_{0}}$

(A) $\frac{2 m v_{0}}{q E_{0}}$

(B) $\frac{\sqrt{2} m v_{0}}{q E_{0}}$

(D) $\frac{3 m v_{0}}{q E_{0}}$

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Magnetic Effects of Current

A particle of mass

m

and charge

q

has an initial velocity

v = v_{0} \hat{j}

. If an electric field

E = E_{0} \hat{i}

and magnetic field

B = B_{0} \hat{i}

act on the particle, its speed will double after a time:

Answer: (C)

The magnetic field points along

x

, the same direction as the electric field. Because

B

lies along

x

, the magnetic force can have no

x

-component, so along

x

the particle feels only the electric force and accelerates uniformly,

v_{x} = \frac{q E _{0}}{m} t

. In the

y

z

plane the magnetic force merely turns the velocity without doing any work, so the speed in that plane stays at its starting value

v_{0}

. The total speed is thus

v = v_{x}^{2} + v_{0}^{2}

, and demanding

v = 2 v_{0}

gives

4 v_{0}^{2} = (\frac{q E _{0} t}{m})^{2} + v_{0}^{2}

, so

(\frac{q E _{0} t}{m})^{2} = 3 v_{0}^{2}

and

t = \frac{3 m v _{0}}{q E _{0}}

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option