Question

A proton goes undflected in a crossed and magnetic field (the fields are perpendicular to each other) at a speed of `2.0X10^5 ms^(-1).` The velocity is perpendicular to both the fields. When the electric field is switched off, the proton moves along a circle of radius 4.0 cm. Find the magnitudes of the electric and the magnetic fields. take the mass of the proton `=1.6 X10^(-27)` kg.
A. `10^(4) N//C, 0.05`
B. `10^(5) N//C, 0.05`
C. `10^(-4) N//C, 0.04`
D. `10^(4) N//C, 0.04`

Answer 1

Correct Answer - A

`F_(m) = Bqv_(0)`, upward direction
`F_(e) = qE`, downward direction
`F_(m) = F_(e) rArr Bv_(0) = E` …(i)
`R = (mv_(0))/(Bq)` …(ii)
`4 xx 10^(-2) = (1.6 xx 10^(-27) xx 2 xx 10^(5))/(B xx 1.6 xx 10^(-19)) rArr B = 0.05 T`
`E = Bv_(0) = 0.05 xx 2 xx 10^(5) = 10^(4) N//C`