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(D) The time period of a charged particle in a magnetic field is given by $T = \frac{2 \pi m}{q B}$. Since the particles are identical ($m$ and $q$ ar

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Both $(A)$ and $(B)$

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(D) The time period of a charged particle in a magnetic field is given by $T = \frac{2 \pi m}{q B}$. Since the particles are identical ($m$ and $q$ are same) and the magnetic field $B$ is uniform,the time period is independent of the angle of entry. Thus,$a = \frac{T_1}{T_2} = 1$.The radius of the helical path is given by $r = \frac{m v \sin 	heta}{q B}$. The ratio of radii is $b = \frac{r_1}{r_2} = \frac{\sin 30^{\circ}}{\sin 60^{\circ}} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$.The pitch of the helical path is given by $p = v \cos 	heta \cdot T$. The ratio of pitches is $c = \frac{p_1}{p_2} = \frac{\cos 30^{\circ}}{\cos 60^{\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.Now,calculating the product $abc = 1 	imes \frac{1}{\sqrt{3}} 	imes \sqrt{3} = 1$. Also,checking option $(B)$: $a = 1$ and $bc = \frac{1}{\sqrt{3}} 	imes \sqrt{3} = 1$. Since $a = bc$ is also true,the correct answer is $(D)$.

Two identical charged particles enter a uniform magnetic field with the same speed but at angles $30^{\circ}$ and $60^{\circ}$ with the field. Let $a, b,$ and $c$ be the ratios of their time periods,radii,and pitches of the helical paths,respectively. Then:

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