Question

An electron, moving along the x-axis with an initial energy of 100 eV, enters a region of magnetic field ⃗B = (1.5 × 10⁻³ T)kˆ 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 electron = 9.1 × 10⁻³¹ kg)
(A) 12.87 cm (B) 2.25 cm
(C) 1.22 cm (D) 11.65 cm

Answer 1

The distance d between P and Q (on the screen) is 12.87cm

Let R be the radius of circular path the electron takes inside the magnetic field, B.

• R = mv/qB = $\sqrt{2m KE}$ / qB
• m =  mass of electron = 9.1 × 10⁻³¹ kg
• K.E =  energy of elextron = 100 eV
• q = charge = 1.6 × 10⁻¹⁹ C
• B = magnetic field = 1.5 × 10⁻³ T

Therefore,

• R = $\sqrt{2 * 9.1 * 10^{-31} * 100 * 1.6 * 10^{-19} }$/$1.6*10^{-19} * 1.5 * 10^{-3}$ = 2.248cm

From the attachment, consider the angle x the electron path makes with the horizontal,

• Inside the field, sin x ≅ 2 / 2 . 248

Therefore

• tan x = 2 / 1 . 026 = QT / U = QT / 6
• = = > QT = 11.69cm

To find PT,

Consider ΔABC,

• From the field, AB = Rcos x
• AS = R
• AS - AB = BS = PT = R - Rcos x = R ( 1 - cos x ) = 2.248( 1 - 1.025/2.248)
• PT = 1.22cm

Therefore distance d between P and Q is 11.69 + 1.22 = 12.91 cm

Closest answer is 12.87cm.