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.

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written 8.2 years ago by

zuzu.zz299 • 0

modified 4.1 years ago by

krithikkm200 • 10

Subject: Applied Physics 2

Topic: Motion Of Charged Paqrticle In Electric And Magnetic Field

Difficulty: Medium

engineering physics

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1 Answer

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written 8.1 years ago by

zuzu.zz299 • 0

modified 8.1 years ago by

infogazi16 • 0

$$ $$ $ 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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