JEE Main 2007
LEVELJEE Advanced
Visualized Solution
The Sigma Insight: Motion of a Charge in Magnetic Fields
The Magic of Crossed Fields
Imagine a charged particle zooming through space. Suddenly, it enters a region where an electric field and a magnetic field are locked in a perpendicular embrace. This is what physicists call a "crossed field" setup.
In this specific problem, the particle's velocity is also perpendicular to both fields. It's a perfect 3D dance: velocity along the x-axis, electric field along the y-axis, and magnetic field along the z-axis.
The Master Equation
Lorentz Force
The problem gives us a massive clue: the particle emerges without any change in its velocity. Constant velocity means zero acceleration. By Newton's second law, the net force acting on the particle must be exactly zero.
The total force on a moving charge is governed by the Lorentz force law. It states that the net force is the vector sum of the electric force and the magnetic force.
Algebraic Manipulation
Since the charge is non-zero, we can divide the entire equation by . This simplifies our life immensely.
Rearranging this, we find that the electric field must perfectly balance the magnetic force term.
The Vector Triple Product
Now comes the mathematical trick. We need to isolate the velocity vector . To do this, we take the cross product of both sides with the magnetic field vector from the right.
This looks intimidating, but we have a powerful tool: the vector triple product identity.
Applying this to our equation, the right side expands beautifully.
Final Calculation
Here is where the geometry of the problem saves the day. The velocity is perpendicular to the magnetic field . Therefore, their dot product is exactly zero.
Furthermore, the dot product of any vector with itself is just the square of its magnitude.
Substituting these back into our expanded equation, the first term vanishes, and the negative signs cancel out.
Dividing by , we arrive at our final, elegant expression for the velocity.
This result is the fundamental operating principle of a velocity selector, a device used in mass spectrometers to filter particles based on their speed!
Similar Questions
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A particle of charge and mass moving under the influence of a uniform electric field and uniform magnetic field follows a trajectory from to as shown in figure. The velocities at and are and . Which of the following statement(s) is/are correct ?
* Multiple Correct Options
(A)
(B)
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(C)
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(D)
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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)
LEVELJEE Advanced
For a positively charged particle moving in a plane initially along the -axis, there is a sudden change in its path due to the presence of electric and/or magnetic fields beyond . The curved path is shown in the plane and is found to be non-circular.
(A)
(B)
(C)
(D)