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(A) When a charged particle moves in a magnetic field with a velocity component $v_{\perp}$ perpendicular to the field and $v_{\parallel}$ parallel to

Vedclass · Accepted Answer

$\frac{2 \pi m v}{q B}$

Vedclass · Answer

(A) When a charged particle moves in a magnetic field with a velocity component $v_{\perp}$ perpendicular to the field and $v_{\parallel}$ parallel to the field,it follows a helical path.The time period $T$ for one complete rotation is determined by the perpendicular component of velocity:$T = \frac{2 \pi m}{q B}$The distance moved along the magnetic field in one rotation is called the pitch $(p)$.$p = v_{\parallel} 	imes T$Substituting the value of $T$:$p = v_{\parallel} 	imes \frac{2 \pi m}{q B}$If the total velocity $v$ is considered as the parallel component,the distance is $\frac{2 \pi m v}{q B}$.

The distance moved by a charged particle along the magnetic field (the component of velocity is parallel to the magnetic field) in one rotation is given by ($m$ - mass of the particle,$v$ - velocity of the particle,$q$ - charge of the particle,$B$ - magnetic field).

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