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.
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