Two charged particles traverse identical heli | vedclass

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(D) The radius of a helical path is given by $R = \frac{mv_{\perp}}{qB}$ and the pitch is $p = \frac{2\pi m v_{\parallel}}{qB}$.Since the helical path

Vedclass · Accepted Answer

The charge to mass ratio satisfies $(\frac{e}{m})_1 + (\frac{e}{m})_2 = 0$

Vedclass · Answer

(D) The radius of a helical path is given by $R = \frac{mv_{\perp}}{qB}$ and the pitch is $p = \frac{2\pi m v_{\parallel}}{qB}$.Since the helical paths are identical,both the radius $R$ and the pitch $p$ must be the same for both particles.This implies that the magnitude of the ratio $|\frac{q}{m}|$ must be the same for both particles.Since the particles traverse the paths in a completely opposite sense,one particle must be positively charged and the other negatively charged.Therefore,$\frac{q_1}{m_1} = -\frac{q_2}{m_2}$,which leads to $(\frac{q}{m})_1 + (\frac{q}{m})_2 = 0$.

Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field $B = B_0 \hat{k}$.

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