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An electron moves in a uniform magnetic field of 0.23 $Wb/m^2$ making an angle of $45^0$ with it. Determine radius & pitch of its helical path. Assuming electron speed to be $3 \times 10^7$ m/sec.
written 6.3 years ago by gravatar for neeta.vanage neeta.vanage • 860 modified 2.2 years ago by gravatar for pedsangini276 pedsangini276 • 4.7k

Subject: Applied Physics 2

Topic: Motion Of Charged Paqrticle In Electric And Magnetic Field

Difficulty: Medium
engineering physics
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written 6.2 years ago by gravatar for neeta.vanage neeta.vanage • 860 modified 6.2 years ago by gravatar for saurabhsingh saurabhsingh • 40

$$ $$ $ B = 0.23 Wb/ m2 , \\ R = ? \\ \text{Pitch of the helix } = ? \\ ϴ = 45° \\ v = 3* 10^7 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} * 3* 10^7 * sin 45 / 0.23 *1.6 *10^{-19} \\ R = 52.4 mm \\ \text{Pitch of the helix is P = 2П m v cos ϴ / Be } \\ P = 2П*9.1 *10^{-31} *3 * 10^7* cos 45 / 0.23 * 1.6 *10^-19 \\ P = 3.29 mm \\ $
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