An electron, moving along the x-axis with an initial energy of 100 eV, enters a region of magnetic field \(\vec B = \left( {1.5 \times {{10}^{ - 3}}T} \right)\hat k\) at S (see figure). The field extends between x = 0 and x = 2 cm. The electron is detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P and Q (on the screen) is:

(Electron’s charge = 1.6 × 10^{-19} C; Mass of the electron = 9.1 × 10^{-31} kg

This question was previously asked in

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

Option 2 : 12.87 cm

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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90 Questions
360 Marks
180 Mins

**Concept:**

The elaborated diagram for the given figure is:

Let’s find value of ‘R’:

In ⊿CBA:

\(\sin {\rm{\theta }} = \frac{x}{R}\)

Where,

x = 8 cm - 6 cm = 2 cm

Now,

\( \Rightarrow \sin {\rm{\theta }} = \frac{2}{R}\)

Now, the radius ‘R’ of an electron with mass ‘m’ moving in magnetic field is given by the formula:

\(R = \frac{{mv}}{{qB}} = \frac{p}{{qB}}\)

Where,

mv = p = Momentum of the electron

The momentum of kinetic energy is related by the formula:

\( \Rightarrow {E_k} = \frac{{{p^2}}}{{2m}}\)

⇒ p^{2 }= 2mE_{k}

\(\therefore p = \sqrt {2m{E_k}} \)

Now,

\( \Rightarrow R = \frac{{\sqrt {2m{E_k}} }}{{qB}}\)

Where,

m is the mass of the electron = 9.1 × 10^{-31} kg (given)

E_{k} is the kinetic energy = 100 eV = 100 × 1.6 × 10^{-19} J = 1.6 × 10^{-17} J (given)

q is the charge of the electron = 1.6 × 10^{-19} C (given)

B is the magnetic field = 1.5 × 10^{-3} T (given)

**Calculation:**

On substituting the values,

\( \Rightarrow R = \frac{{\sqrt {\left( {2 \times 9.1 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 17}}} \right)} }}{{1.6 \times {{10}^{ - 19}} \times 1.5 \times {{10}^{ - 3}}}}\)

∴ R = 2.24 cm

Now,

\( \Rightarrow \sin {\rm{\theta }} = \frac{2}{{2.24}}\)

∴ θ = 63.23°

Let’s find value of ‘h’:

In ⊿EAQ:

\( \Rightarrow \tan {\rm{\theta }} = \frac{h}{6}\)

\( \Rightarrow 1.982 = \frac{h}{6}\)

∴ h = 11.82

From diagram,

⇒ d = h + R(1 - cos θ)

⇒ d = 11.82 + 2(1 - 0.45)

⇒ d = 11.82 + 2(0.55)

⇒ d = 11.82 + 1.099

∴ d = 12.91 cm ≈ 12.87 cm