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28. An electron with a speed of 4.0 x 10° m/s enters a uniform magnetic field of magnitude 0.040 T at an angle of 35 degrees to the magnetic field lines. The electron will follow a helical path. a) Determine the radius of the helical path. b) How far forward will the electron have moved after completing one circle?​

Sagot :

The radius of the helical path and the period of the electron are 3. 26* 10^-10m and 8.94*10^-10m respectively.

How to determine the radius

Using the formula:

r = [tex]\frac{mvsin \alpha }{Bq}[/tex]

m = 9.109 × 10-31 kg

α = 35°

B = 0. 040 T

q = 1.6 × 10-19 C

Substitute values into the formula

[tex]r = \frac{9. 109 * 10^-31 * 4.0 * 10^0 * sin 35}{0. 040 * 1.6 * 10^-19 }[/tex]

[tex]r = \frac{2. 089 * 10^-30}{6. 4* 10^-21}[/tex]

[tex]r = 3. 26 * 10^-10[/tex] m

b. [tex]T = \frac{2\pi m}{qB}[/tex]

[tex]T = \frac{2 * 3.142 * 9.109 *10^-31}{1.6* 10^-19* 0.040}[/tex]

[tex]T = \frac{5. 72* 10^-30}{6. 4* 10^-21}[/tex]

[tex]T = 8. 94* 10^-10[/tex]

Therefore, the radius of the helical path and the period of the electron are 3. 26* 10^-10m and 8.94*10^-10m respectively.

Learn more about magnetic field here:

https://brainly.com/question/7802337

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