Question

A charge particle having mass of 1.6 × 10^–26 kg comes out of accelerator tube with kinetic energy2×10³ eV. Calculate the smallest magnitude of magnetic field that should be applied in vertically downwards direction to just prevent the charge particle from colliding the plate (Assume charge on particle = charge of proton)

• A. 2 Tesla
• B. 4 Tesla
• C. .02 Tesla
• D. 0.04 Tesla

Answer 1

Answer: (A)

2 Tesla

Answer 2

r = d or = d

Ž B =

or B = $\sqrt { \frac { 2 \times 1.6 \times 10 ^ { - 26 } \times 2 \times 10 ^ { 3 } \times \mathrm { e } } { \mathrm { e } ^ { 2 } \times 10 ^ { - 4 } } }$

B = $\sqrt { \frac { 2 \times 1.6 \times 10 ^ { - 26 } \times 2 \times 10 ^ { 3 } } { 1.6 \times 10 ^ { - 19 } \times 10 ^ { - 4 } } }$= 2 Tesla