Consider the motion of a positive point charge in a region, there are simultaneous uniform electric and magnetic fields {E} = {E}₀j} and {B} = {B}₀j}. At time t = 0, this charge has velocity {v} in the xy-plane, making an angle θ with the x-axis. Which of the following option(s) is(are) correct for time t > 0? [JEE Adv 2012 P1]

If θ = 0°, the change moves in a circular path in the xy-plane.

If θ = 0°, the change undergoes helical motion with constant pitch along the y-axis.

If θ = 10°, the change undergoes helical motion with its pitch increasing with time, along the y-axis.

If θ = 90°, the change undergoes linear but accelerated motion along the y-axis.

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Magnetism

Section: Physics

Question : 36 of 46

Marks: +1, -0

Consider the motion of a positive point charge in a region, there are simultaneous uniform electric and magnetic fields

\vec{E}

\hat{E}_{0j}

and

\vec{B}

\vec{B}_{0j}

. At time t = 0, this charge has velocity

\vec{v}

in the xy-plane, making an angle θ with the x-axis. Which of the following option(s) is(are) correct for time t > 0?

[JEE Adv 2012 P1]

If θ = 0°, the change moves in a circular path in the xy-plane.
If θ = 0°, the change undergoes helical motion with constant pitch along the y-axis.
If θ = 10°, the change undergoes helical motion with its pitch increasing with time, along the y-axis.
If θ = 90°, the change undergoes linear but accelerated motion along the y-axis.

Validate

Solution:

• For θ = 90°:

\vec{v}

\vec{E}

and

\vec{v}

\vec{B}

. Therefore, magnetic field

\vec{B}

exerts no force on the point charge:

\vec{F}_B

q(\vec{v} \times \vec{B})

= 0

\vec{F}_E

\vec{qE}

Therefore, the motion is along y-axis and the particle accelerates.

• For θ = 0° and θ = 10°: The motion is helical with increasing pitch, along with increasing time (along y-axis), as p = (vcosq)t, where v increases due to

\vec{F}_E

. (For θ = 0°, the charge cannot move in a circular path.)

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