Sigma Percentile
LEVELJEE Main

Animated Solution for Physics - Magnetic Effects of Current: A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity, then

Select Answer:

Visualized Solution

Visualizing the Setup

  • Given:
  • An electron () is projected parallel to both fields.

The Lorentz Force

  • The total force on a charged particle in electromagnetic fields is given by the Lorentz Force:

Analyzing the Magnetic Force

  • Magnetic Force:
  • Since , the angle .

Analyzing the Electric Force

  • Electric Force:
  • For an electron, .
  • The negative sign indicates the force is opposite to the electric field.

Conclusion

  • Net Force:
  • Since the force is opposite to the velocity , it causes deceleration.
  • Result: The velocity of the electron will decrease.

The Way Forward

  • What if the particle was a proton ()?
  • The electric force would be in the same direction as velocity, causing it to accelerate.

The Sigma Insight: Motion of a Charge in Magnetic Fields

Solution Diagram

The Setup

A Tale of Two Fields
Imagine a region of space where both an electric field and a magnetic field are perfectly aligned, pointing in the exact same direction. Into this highly structured environment, we fire an electron. We shoot it straight along the field lines, meaning its velocity vector is parallel to both and .
To understand the fate of this electron, we must analyze the forces acting upon it. Will it curve? Will it speed up? Will it slow down? The answer lies in the fundamental laws of electromagnetism.

The Master Equation

Lorentz Force
Whenever a charged particle moves through a region containing both electric and magnetic fields, the total force it experiences is governed by the Lorentz force equation:
This elegant equation is actually a combination of two distinct forces: the electric force and the magnetic force . Let's break them down one by one to see how they influence our electron.

The Magnetic Force

A Cross Product Mystery
Let's look at the magnetic component first. The magnetic force depends on the cross product of the velocity vector and the magnetic field vector. The magnitude of this force is given by:
where is the angle between and . In our specific scenario, the electron is projected along the direction of the magnetic field. This means the angle is exactly .
Since , the entire cross product collapses to zero. The magnetic field, despite being present, exerts absolutely no force on the electron. The electron is completely blind to the magnetic field as long as it moves perfectly parallel to it.

The Electric Force

The Power of a Negative Charge
Now, let's turn our attention to the electric force. The electric force is much simpler; it doesn't care about the particle's velocity or direction of motion. It only cares about the charge and the electric field:
Here is where the crucial detail comes in: our particle is an electron. An electron carries a negative charge, so . Substituting this into our equation gives:
That negative sign is everything. It tells us that the electric force acts in the exact opposite direction of the electric field. Since the electron was moving in the direction of the electric field, the force is pushing against it, directly opposing its motion.

The Final Verdict

Deceleration
We have established that the magnetic force is zero and the electric force is pushing backward against the electron's velocity.
When a net force acts in the direction opposite to an object's motion, it causes deceleration. The electron will continue to move in a straight line (because there are no sideways forces to turn it), but it will progressively lose speed. Therefore, the correct conclusion is that its velocity will decrease.

Similar Questions

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An electron and a proton are moving on straight parallel paths with same velocity. They enter a semi-infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement(s) is/are true?

* Multiple Correct Options
(A)
They will never come out of the magnetic field region
(B)
They will come out travelling along parallel paths
(C)
They will come out at the same time
(D)
They will come out at different times
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An electron moving with a speed along the positive -axis at enters a region of uniform magnetic field which exists to the right of -axis. The electron exits from the region after sometime with the speed at coordinate , then

(A)
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In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. A charged particle is released from rest in this region. The path of the particle will be a

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Comprehension Passage

A charged particle (electron or proton) is introduced at the origin () with a given initial velocity . A uniform electric field and a uniform magnetic field exist everywhere. The velocity , electric field and magnetic field are given in columns 1, 2 and 3, respectively. The quantities are positive in magnitude. $\begin{array}{lll} \hline \text{Column 1} & \text{Column 2} & \text{Column 3} \\ \hline \text{(I) Electron with } \mathbf{v} = 2\frac{E_0}{B_0}\hat{x} & \text{(i) } \mathbf{E} = E_0\hat{z} & \text{(P) } \mathbf{B} = -B_0\hat{x} \\ \text{(II) Electron with } \mathbf{v} = \frac{E_0}{B_0}\hat{y} & \text{(ii) } \mathbf{E} = -E_0\hat{y} & \text{(Q) } \mathbf{B} = B_0\hat{x} \\ \text{(III) Proton with } \mathbf{v} = 0 & \text{(iii) } \mathbf{E} = -E_0\hat{x} & \text{(R) } \mathbf{B} = B_0\hat{y} \\ \text{(IV) Proton with } \mathbf{v} = 2\frac{E_0}{B_0}\hat{x} & \text{(iv) } \mathbf{E} = E_0\hat{x} & \text{(S) } \mathbf{B} = B_0\hat{z} \\ \hline \end{array}$
Question 1:

In which case would the particle move in a straight line along the negative direction of Y-axis (i.e. move along )?

(A)
(IV) (ii) (S)
(B)
(II) (iii) (Q)
(C)
(III) (ii) (R)
(D)
(III) (ii) (P)
Question 2:

In which case will the particle move in a straight line with constant velocity?

(A)
(II) (iii) (S)
(B)
(III) (iii) (P)
(C)
(IV) (i) (S)
(D)
(III) (ii) (R)
Question 3:

In which case will the particle describe a helical path with axis along the positive z-direction?

(A)
(II) (ii) (R)
(B)
(III) (iii) (P)
(C)
(IV) (i) (S)
(D)
(IV) (ii) (R)
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If an electron and a proton having same momenta enter perpendicularly to a magnetic field, then

(A)
curved path of electron and proton will be same (ignoring the sense of revolution)
(B)
they will move undeflected
(C)
curved path of electron is more curved than that of proton
(D)
path of proton is more curved
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An electron is moving along +x-direction with a velocity of . It enters a region of uniform electric field of pointing along +y-direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the x-direction will be

(A)
, along + z-direction
(B)
, along − z-direction
(C)
, along + z-direction
(D)
, along − z-direction
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A charged particle with charge enters a region of constant, uniform and mutually orthogonal fields and with a velocity perpendicular to both and and comes out without any change in magnitude or direction of . Then,

(A)
(B)
(C)
(D)
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A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a

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cycloid
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A charged particle moves through a magnetic field perpendicular to its direction. Then,

(A)
the momentum changes but the kinetic energy is constant
(B)
both momentum and kinetic energy of the particle are not constant
(C)
both momentum and kinetic energy of the particle are constant
(D)
kinetic energy changes but the momentum is constant
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Two particles and of masses and respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are and , respectively and the trajectories are as shown in the figure. Then

(A)
(B)
(C)
and
(D)
and