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
[tex]r = \dfrac{mv}{qB}[/tex]
m is mass of electron
q is charge of electron
[tex]r = \dfrac{9.11 \times 10^{-31}\times 7.50\times 10^6}{1.6\times 10^{-19}\times 1 \times 10^{-5}}[/tex]
r = 4.266 m
Hence, radius of circular path is equal to 4.266 m.
The radius of the circular path the electron follows will be "4.266 m". To understand the calculation check below.
According to the question,
Electron's speed = 7.50 × 10⁶ m/s
Magnetic field, B = 1 × 10⁻⁵ T
Electron's mass. m = 9.11 × 10⁻³¹
Charge of electron, q = 1.6 × 10⁻¹⁹
We know the relation of radius of curvature,
→ r = [tex]\frac{9.11\times 10^{-31}\times 7.50\times 10^6}{1.6\times 10^{-19}\times 1\times 10^{-5}}[/tex]
= 4.266 m
Thus the above answer is right.
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