An electron moving along the X-axis with an initial energy of 100 eV, enters a region of magnetic fi

Q: Why can't I just use the standard formula $d = \frac{qBLD}{mv}$?

That formula assumes the angle of deflection is very small ($\tan\theta \approx \sin\theta \approx \theta$). Here, the radius ($2.25$ cm) is very close to the field width ($2$ cm), resulting in a huge deflection angle ($\approx 63^\circ$). The approximation completely breaks down.

Q: How did you know to use $K = 100$ eV in the momentum formula?

The kinetic energy $K$ of the electron is given as $100$ eV. The relationship between momentum $p$ and kinetic energy $K$ is $p = \sqrt{2mK}$. Remember to convert eV to Joules by multiplying by $1.6 \times 10^{-19}$.

Q: Where does the formula $y_1 = R - \sqrt{R^2 - L^2}$ come from?

If you draw a circle of radius $R$ with its center at $(0, R)$, its equation is $x^2 + (y-R)^2 = R^2$. Solving for $y$ when $x = L$ gives the vertical position $y_1 = R - \sqrt{R^2 - L^2}$.

Q: Why does the electron travel in a straight line after $x=2$ cm?

The magnetic field only exists between $x=0$ and $x=2$ cm. Once the electron exits this region, there is no net force acting on it, so Newton's First Law dictates it must travel in a straight line at a constant velocity.

Q: Why did the official solution key say 'No option is matching'?

The authors of the official solution likely fell into the very trap they set! They used the small-angle approximation formula, got $5.34$ cm, and concluded no option matched. However, a rigorous geometric calculation yields $12.87$ cm, which is exactly option (b).

Question

An electron moving along the X-axis with an initial energy of $100 	ext{ eV}$, enters a region of magnetic field $\mathbf{B} = (1.5 	imes 10^{-3} 	ext{ T})\hat{\mathbf{k}}$ at $S$ (see figure). The field extends between $x = 0$ and $x = 2 	ext{ cm}$. The electron is detected at the point $Q$ on a screen placed $8 	ext{ cm}$ away from the point $S$. The distance $d$ between $P$ and $Q$ (on the screen) is
(Take, electron's charge $= 1.6 	imes 10^{-19} 	ext{ C}$, mass of electron $= 9.1 	imes 10^{-31} 	ext{ kg}$)

Options:

(A) 11.65 cm

(B) 12.87 cm

(C) 1.22 cm

(D) 2.25 cm

Sigma Percentile · Accepted Answer

### The Trap of the Small Angle Approximation

Welcome to a classic JEE trap! At first glance, this looks like a standard magnetic deflection problem. An electron enters a magnetic field, curves a bit, and hits a screen. Many students immediately reach for the standard small-angle approximation formula: $d \approx \frac{qBLD}{mv}$.

However, if you blindly use this formula here, you will get an answer around $5.34 	ext{ cm}$, which is nowhere to be found in the options! The official solution key even claimed "No option is matching" because the authors themselves fell into this very trap. Let's uncover the beautiful geometry that leads to the correct answer.

### Phase 1: The Radius Reveal

First, we must determine the exact radius $R$ of the electron's circular path inside the magnetic field. The magnetic Lorentz force provides the necessary centripetal force:

$$R = \frac{mv}{qB}$$

Since we are given the kinetic energy $K = 100 	ext{ eV}$, we can express the momentum $p = mv$ as $\sqrt{2mK}$. Substituting the given values (and remembering to convert eV to Joules):

$$R = \frac{\sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 100 	imes 1.6 	imes 10^{-19}}}{1.6 	imes 10^{-19} 	imes 1.5 	imes 10^{-3}}$$

$$R = \frac{\sqrt{29.12 	imes 10^{-48}}}{2.4 	imes 10^{-22}} \approx \frac{5.4 	imes 10^{-24}}{2.4 	imes 10^{-22}} = 2.25 	ext{ cm}$$

**Here is the critical realization:** The width of the magnetic field is $L = 2 	ext{ cm}$. The radius of the path is $R = 2.25 	ext{ cm}$. The radius is barely larger than the field width! This means the electron will undergo a massive deflection angle (nearly $63^\circ$). The small-angle approximation $	an	heta \approx \sin	heta \approx 	heta$ is completely invalid here.

### Phase 2: Inside the Magnetic Field

We must trace the exact geometric path. Inside the field, the electron travels along a circular arc. Let's calculate its vertical deflection, $y_1$, at the exact moment it exits the field at $x = L = 2 	ext{ cm}$. From the geometry of a circle centered at $(0, R)$:

$$y_1 = R - \sqrt{R^2 - L^2}$$

$$y_1 = 2.25 - \sqrt{2.25^2 - 2^2} = 2.25 - \sqrt{5.0625 - 4}$$

$$y_1 = 2.25 - \sqrt{1.0625} \approx 2.25 - 1.03 = 1.22 	ext{ cm}$$

### Phase 3: The Straight Line Dash

Once the electron leaves the magnetic field, no forces act upon it. Newton's First Law dictates it will travel in a straight line towards the screen. To find its path, we need the slope of the trajectory at the exit point, which is $	an	heta$:

$$	an	heta = \frac{L}{\sqrt{R^2 - L^2}} = \frac{2}{1.03} \approx 1.94$$

The remaining horizontal distance to the screen is $D - L = 8 - 2 = 6 	ext{ cm}$. In this straight-line path, the additional vertical deflection $y_2$ is simply the horizontal distance multiplied by the slope:

$$y_2 = 6 	imes 	an	heta = 6 	imes 1.94 = 11.64 	ext{ cm}$$

### Final Calculation

The total deflection $d$ on the screen is the sum of the deflection inside the field and the deflection outside the field:

$$d = y_1 + y_2 = 1.22 + 11.64 = 12.86 	ext{ cm}$$

Accounting for slight rounding differences in the constants, this perfectly matches **Option (b): 12.87 cm**. This problem is a masterclass in why you should always verify the physical constraints of a system before blindly applying approximation formulas!

Select Answer:

Visualized Solution

Analyzing the Setup

Radius of Circular Path

Substituting Values

Calculating the Radius

Deflection Inside the Field ( $y_{1}$ )

Calculating $y_{1}$

Exit Angle ( $θ$ )

Path Outside the Field

Deflection Outside the Field ( $y_{2}$ )

Total Deflection ( $d$ )

The Sigma Insight: Motion of a Charge in Magnetic Fields

The Trap of the Small Angle Approximation

Phase 1

Phase 2

Phase 3

Final Calculation

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