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A proton describes a circle of radius $1 cm$ in a magnetic field of strength $0 .10 T$ . What would be the radius (in cm) of the circle described by an α -particle moving with the same speed in the same magnetic field?

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detailed solution

Correct option is B

$\begin{array}{l} r = \frac{mv}{qB} \\ 0 .01 = \frac{mv}{e 0 .1} \dots \dots (1) \\ r' = \frac{4 m \times v}{2 e \times 0 .1} \dots \dots (2) \\ (2) \div (1) \\ \Rightarrow \frac{r'}{0 .01} = \frac{4 mve \times 0 .1}{2 e \times 0 .1 \times mv} = \frac{4}{2} = 2 \\ \Rightarrow r' = 0 .02 m = 2 cm . \end{array}$

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detailed solution

Correct answer is 2

$\begin{array}{l} r = \frac{mv}{qB} \\ 0 .01 = \frac{mv}{e 0 .1} \dots \dots (1) \\ r' = \frac{4 m \times v}{2 e \times 0 .1} \dots \dots (2) \\ (2) \div (1) \\ \Rightarrow \frac{r'}{0 .01} = \frac{4 mve \times 0 .1}{2 e \times 0 .1 \times mv} = \frac{4}{2} = 2 \\ \Rightarrow r' = 0 .02 m = 2 cm . \end{array}$

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