An electron of mass m and charge q is travelling with a speed v along a circular path of radius r at right angles to a uniform magnetic field B. If speed of the electron is doubled and the magnetic field is halved, then resulting path would have a radius of

$\frac{r}{4}$

$\frac{r}{2}$

2r

4r

Solution

In a perpendicular magnetic field

Magnetic force = centripetal force

Bqv = $\frac{{mv}^{2}}{r}$

⇒ r = $\frac{mv}{Bq}$

∴ $\frac{r_{1}}{r_{2}} = \frac{v_{1}}{v_{2}} \times \frac{B_{2}}{B_{1}}$

$\frac{r_{1}}{r_{2}} = \frac{1}{2} \times \frac{1}{2}$

$r_{2} = 4 r_{1}$

⇒ $r_{2} = 4 r$

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