Two charged particles traverse identical heli | vedclass

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(D) The radius of a helical path is given by $R = \frac{mv_{\perp}}{qB}$. Since the paths are identical,$R_1 = R_2$,which implies $\frac{m_1 v_{\perp

Vedclass · Accepted Answer

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

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(D) The radius of a helical path is given by $R = \frac{mv_{\perp}}{qB}$. Since the paths are identical,$R_1 = R_2$,which implies $\frac{m_1 v_{\perp 1}}{q_1 B} = \frac{m_2 v_{\perp 2}}{q_2 B}$.Given that the particles traverse the paths in opposite senses,the charges must have opposite signs,i.e.,$q_1 = -q_2$.For the helical paths to be identical,the pitch $p = \frac{2\pi m v_{\parallel}}{qB}$ must also be the same. Since the sense of rotation is opposite,the angular frequency $\omega = \frac{qB}{m}$ must have opposite signs.If the paths are identical in geometry and pitch,the magnitude of the charge-to-mass ratio must be equal,but the signs must be opposite.Thus,$\frac{q_1}{m_1} = -\frac{q_2}{m_2}$,which leads to $\frac{q_1}{m_1} + \frac{q_2}{m_2} = 0$.

Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field $B = B_0 \hat{k}$. Which of the following statements is true?

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