Physics: A cosmic ray electron moves at 7.50×106 m/s perpendicular to the Earth’s magnetic field at an altitude where field strength is 1.00×10−5 T . What is the radius of the circular path the electron follows?

A cosmic ray electron moves at 7.50×106 m/s perpendicular

Ответ:

mnrtunes

21.01.2022

Physics

4.266 m.

Explanation:

Given,

Speed of the electron = 7.50 x 10⁶ m/s

Magnetic field, B = 1 x 10⁻⁵ T

radius of circular path = ?

Using formula of radius of curvature of the charge particle

$r = \dfrac{mv}{qB}$

m is mass of electron

q is charge of electron

$r = \dfrac{9.11 \times 10^{-31}\times 7.50\times 10^6}{1.6\times 10^{-19}\times 1 \times 10^{-5}}$

r = 4.266 m

Hence, radius of circular path is equal to 4.266 m.

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Ответ:

mkim2290

26.08.2020

Physics

0.07 rad/s

Explanation:

Given

kite is 100 ft high

and moves horizontally at 7 ft/s

Total string let out =200 ft

String length(l), vertical(y) & Horizontal(x) distance of kite will form a right angle triangle

L^2=y^2+x^2

Differentiating both side

$0=2y\frac{\mathrm{d} y}{\mathrm{d} t}+2x\frac{\mathrm{d} x}{\mathrm{d} t}$

$y\frac{\mathrm{d} y}{\mathrm{d} t}=-x\frac{\mathrm{d} x}{\mathrm{d} t}$

$100\times \frac{\mathrm{d} y}{\mathrm{d} t}=-\sqrt{3}\times 100\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} y}{\mathrm{d} t}=7\sqrt{3}$

Now $Lcos\theta =x$

Differentiating

$Lsin\theta \times \frac{\mathrm{d} \theta }{\mathrm{d} t}=\frac{\mathrm{d} x}{\mathrm{d} t}$

$200\times \frac{100}{200}\times \frac{\mathrm{d} \theta }{\mathrm{d} t}=7$

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=0.07 rad/s$

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A cosmic ray electron moves at 7.50×106 m/s perpendicular to the Earth’s magnetic field at an altitude where field strength is 1.00×10−5 T . What is the radius of the circular path the electron follows?

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