Answer to Question #120398 in Other for Nimra

Question

an electron at pointing in the figure has a speed v of 1.41 E 6 m/s find (a )the magnitude and direction of the magnetic field that will cause the electron to flow the semicircular path from A to B and( b) the time required for the electron to move from A to B.

Answer 1

a) At A, magnetic force should be towards right. Thus from Fleming's left hand rule magnetic field should be inwards (into a paper).

The electron experiences centripetal force under the action of magnetic field. Hence

$${m_ev_0^2\over r}=|e|v_0B=>B={m_ev_0\over |e|r}$$

$$B={9.11\times10^{-31}kg\cdot 1.41\times 10^6 m/s\over 1.6\times 10^{-19}C\cdot 0.05m}=1.61\times 10^{-4}T$$

b)

$$t={\pi r\over v_0}={\pi\cdot 0.05 m\over 1.4\times 10^6 m/s}=1.12\times10^{-7}s=0.112 \mu s$$