An electron moves with a velocity 1 × 10³ m/s in a magnetic field of induction 0.3 T at an angle 30°. If e/m of electron is 1.76 × 10¹¹ C/kg, the radius of the path is nearly (a) 10⁻⁸ m (b) 2 × 10⁻⁸ m (c) 10⁻⁶ m (d) 10⁻¹⁰ m

Answer: (a). Only the perpendicular component makes the circle: r = (mvsin θ)/(eB) = (vsin θ)/((e/m)B). r = 10³ × 0.51.76 × 10¹¹ × 0.3 = 5005.28 × 10¹⁰ ≈ 9.5 × 10⁻⁹ m So the answer is (a) 10⁻⁸ m.

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Moving Charge in a Magnetic Field

An electron moves with a velocity $1 \times 10^{3}\text{ m/s}$ in a magnetic field of induction $0.3\text{ T}$ at an angle $30^\circ$ . If $\frac{e}{m}$ of electron is $1.76 \times 10^{11}\text{ C/kg}$ , the radius of the path is nearly

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Answer: (a)

Only the perpendicular component makes the circle: $r = \dfrac{mv\sin\theta}{eB} = \dfrac{v\sin\theta}{(e/m)B}$ .
$r = \dfrac{10^{3} \times 0.5}{1.76 \times 10^{11} \times 0.3} = \dfrac{500}{5.28 \times 10^{10}} \approx 9.5 \times 10^{-9}$ m
So the answer is (a) $10^{-8}\text{ m}$ .

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