An electron, moving along the x-axis with an initial energy of 100eV, enters a region of magnetic field B→=(1.5×10⁻³T)k^ at S (See figure). The field extends between x=0 and x=2cm. The electron is detected at the point Q on a screen placed 8cm away from the point S. The distance d between P and Q (on the screen) is :(electron's charge =1.6×10⁻¹⁹C, mass of electron =9.1×10⁻³¹kg)
12.87cm
11.65cm
1.22cm
2.25cm
Solution:
Correct option is B. 1.22cm R=mv/qB=√(2m(K.E))/qB R=√(2×9.1×10⁻³¹×(100×1.6×10⁻¹⁹))/(1.6×10⁻¹⁹×1.5×10⁻³) R=2.248cm sin θ = 2R/8 tan θ = d/8 QU/TU ≈ 2.026 = QU/8 QU = 11.69cm PU = R(1-cos θ) = 1.22cm d = QU + PU