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Dennis_Churaev [7]

3 years ago

10

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?

1 answer:

wlad13 [49]3 years ago

4 0

Answer:

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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The radius of the pulley is $R = 1.5m$

The radius of the cord around the pulley is $r = 1.5 m$

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Now when the pulley is rotated that torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

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$a = r\alpha$

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$mgr = I\alpha + m(r\alpha ) r$

$mgr = \alpha [ I + mr^2]$

making $\alpha$ the subject

$\alpha = \frac{mgr}{I -mr ^2}$

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$\alpha = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}$

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$w = \frac{8.38}{1.5}$

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$x = \frac{5.59 * 0.159155}{1}$

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Now z is mathematically obtained as

$z = \frac{ 1}{60}$

z $= 0.01667 \ minute$

Therefore

$w = \frac{0.8892}{0.01667}$

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