A charged particle enters a uniform magnetic field with its velocity at an angle of \( 45^\circ \) to the field. The pitch of its helical path is p. The radius of the helix is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \frac{p}{2\pi} \)
The velocity splits into a component \( v_{\perp}=v\sin45^\circ \) across the field and a component \( v_{\parallel}=v\cos45^\circ \) along it. The first gives circular motion of radius \( mv_{\perp}/qB \). The second carries the particle along the field. In one revolution it moves through the pitch \( v_{\parallel}T \), with \( T=2\pi m/qB \).
\( \displaystyle r=\frac{mv\sin45^\circ}{qB} \)
\( \displaystyle p=\frac{2\pi m\,v\cos45^\circ}{qB} \)
Dividing the two:
\( \begin{aligned} \frac{r}{p} &= \frac{\sin45^\circ}{2\pi\cos45^\circ} \\ &= \frac{1}{2\pi} \end{aligned} \)
\( \displaystyle r=\frac{p}{2\pi} \)
NCERT (2026-27):
Motion in a Magnetic Field – “… If there is a component of the velocity parallel to the magnetic field (denoted by v||), it will make the particle move along the field and the path of the particle would be a helical one … The distance moved along the magnetic field in one rotation is called pitch p. …”
\( \displaystyle p=v_{\parallel}T=\frac{2\pi m\,v_{\parallel}}{qB} \qquad [4.6(b)] \)
“The radius of the circular component of motion is called the radius of the helix.”