An electron of mass m and charge q is moving | vedclass

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(B) The radius of a circular path for a charged particle in a magnetic field is given by the formula: $r = \frac{mv}{Bq}$.Here,the mass $m$ and charge

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$4r$

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(B) The radius of a circular path for a charged particle in a magnetic field is given by the formula: $r = \frac{mv}{Bq}$.Here,the mass $m$ and charge $q$ of the electron remain constant.Therefore,the relationship between the variables is $r \propto \frac{v}{B}$.This implies: $\frac{r_2}{r_1} = \frac{v_2}{v_1} \cdot \frac{B_1}{B_2}$.According to the problem,the new speed $v_2 = 2v$ and the new magnetic field $B_2 = \frac{B}{2}$.Substituting these values: $\frac{r_2}{r} = \frac{2v}{v} \cdot \frac{B}{B/2} = 2 \cdot 2 = 4$.Thus,$r_2 = 4r$.Therefore,the correct option is $B$.

An electron of mass $m$ and charge $q$ is moving with a speed $v$ in a circular path of radius $r$ perpendicular to a uniform magnetic field of intensity $B$. If the speed of the electron is doubled and the magnetic field is halved,what will be the radius of the resulting path?

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