A charged particle in a uniform magnetic fiel | vedclass

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(B) The magnetic field is directed along the $z$-axis $(B = B_0 \hat{k})$.The velocity of the particle is $v = 3 \hat{i} + 4 \hat{k} 	ext{ m/s}$.The

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Helical path, $\frac{1}{2} 	ext{ s}$.

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(B) The magnetic field is directed along the $z$-axis $(B = B_0 \hat{k})$.The velocity of the particle is $v = 3 \hat{i} + 4 \hat{k} 	ext{ m/s}$.The velocity component perpendicular to the magnetic field is $v_{\perp} = 3 \hat{i} 	ext{ m/s}$, which causes circular motion in the $x-y$ plane.The velocity component parallel to the magnetic field is $v_{\parallel} = 4 \hat{k} 	ext{ m/s}$, which remains constant as there is no force acting in this direction.Since the particle has both perpendicular and parallel velocity components, the trajectory is a helical path.The distance to be covered along the $z$-axis is $s = 2 	ext{ m}$.Using the formula $s = v_{\parallel} 	imes t$, we get $t = \frac{s}{v_{\parallel}} = \frac{2}{4} = \frac{1}{2} 	ext{ s}$.

$A$ charged particle in a uniform magnetic field $B = B_0 \hat{k}$ starts moving from the origin with velocity $v = 3 \hat{i} + 4 \hat{k} \text{ m/s}$. The trajectory of the particle and the time $t$ at which it reaches $2 \text{ m}$ above the $x-y$ plane are,

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