An electron travels with a velocity $2.5 \times 10^6$ m/sec in vacuum in a uniform magnetic field strength of $0.94 \times 10^{-4} Wb/m^2$ such that its velocity vector makes an angle of $30^0$ with the field direction. Determine the distance covered along the magnetic induction lines in five such revolutions.

Subject: Applied Physics 2

Topic: Motion Of Charged Paqrticle In Electric And Magnetic Field

Difficulty: Medium

$$ $$ $ B = 0.94*10^{-4} Wb/ m2 , \\ R = ? \\ \text{Pitch of the helix = ? } \\ ϴ = 30° \\ v = 2.5*10^6 m/S \\ R = m v sinϴ / Be \\ \text {Mass of electron }m = 9.1*10^{-31} Kg \\ \text {Charge of electron }=1.6*10^{-19} \\ R = 9.1* 10^-{31} * 2.5*10^6 *sin 30 / 0.94*10^{-4} *1.6 *10^{-19} \\ R = 75.6 mm \\ \text {Distance along the magnetic lines is pitch } \\ \text {Pitch of the helix is P = 2П m v cos ϴ / Be } \\ P = 2П*9.1 x 10^{-31} * 2.5*10^6 * cos 30 / 0.94* 10^{-4} *1.6*10^{-19} \\ P = 0.82 m \\ \text {In 5 revolution 5 *0.82 = 4.115 m} \\ $