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VITEEE 2015 Paper

Section: Physics

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Question : 19 of 120

Marks: +1, -0

An electron of mass 9.0 x

10^{-31}

kg under the action of a magnetic field moves in a circle of radius 2 cm at a speed of 3 x

10^6

m/s. If a proton of mass 1.8 x

10^{27}

kg was to move in a circle of same radius in the same magnetic field, then its speed will become

1.5 x $10^3$ m/s
3 x $10^6$ m/s
6 x $10^4$ m/s
2 x $10^6$ m/s

Solution:

Here, the magnetic force (Bqv) will provide the necessary centripetal force

\left(\frac{mv^2}{r}\right)

∴ Bqv =

\frac{mv^2}{r}

⇒ Bqv =

\frac{mv^2}{r}

⇒ Bqv = mv

For electron and proton, the magnetic field B, charge q and radius r, all same

∴ mv = constant

i.e.

m_e v_e

=

m_p v_p

v_p

=

\left(\frac{m_e}{m_p}\right)v_e

=

\left(\frac{9\times10^{-31}}{1.8\times10^{-27}}\right)

3 ×

10^6

= 1.5 ×

10^3

m/s

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