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(C) The angular velocity is given by $\omega = 4 \pi 	imes 10^6 	ext{ rad s}^{-1}$.The velocity component parallel to the magnetic field is $v_{\par

Vedclass · Accepted Answer

$15$

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(C) The angular velocity is given by $\omega = 4 \pi 	imes 10^6 	ext{ rad s}^{-1}$.The velocity component parallel to the magnetic field is $v_{\parallel} = 3 	imes 10^5 	ext{ m s}^{-1}$.The time period of one complete revolution is $T = \frac{2 \pi}{\omega}$.The pitch of the helix is the distance traveled along the magnetic field direction in one time period: $	ext{Pitch} = v_{\parallel} 	imes T$.Substituting the values: $	ext{Pitch} = (3 	imes 10^5) 	imes \left( \frac{2 \pi}{4 \pi 	imes 10^6} ight)$.$	ext{Pitch} = 3 	imes 10^5 	imes \frac{1}{2 	imes 10^6} = \frac{3}{20} 	ext{ m} = 0.15 	ext{ m}$.Converting to centimeters: $0.15 	ext{ m} = 15 	ext{ cm}$.

$A$ charged particle, when entering a uniform magnetic field, moves in a helical path. If its angular velocity is $4 \pi \times 10^6 \text{ rad s}^{-1}$ and its velocity in the direction of the magnetic field is $3 \times 10^5 \text{ m s}^{-1}$, then the pitch of the helix is: (in $\text{ cm}$)

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