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Standard XII

Physics

Properties of Charges

An electron h...

Question

An electron having energy $10 e V$ is circulating in a path of radius $0.1 m$ having a magnetic field of $10^{- 4} T$ . The speed of the electron will be :

2.0 \times (10^{6}) m s^{- 1}

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4.8 \times (10^{6}) m s^{- 1}

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2.0 \times (10^{12}) m s^{- 1}

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4.8 \times (10^{12}) m s^{- 1}

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Solution

The correct option is A $2.0 \times (10^{6}) m s^{- 1}$
Magnetic force provides necessary centripetal force,
$q V B = \frac{m v^{2}}{r}$
$q V B = \frac{2 K E}{r}$
$V = \frac{2 K E}{q B r}$
= $\frac{2 \times 10}{10^{- 4} \times 0.1}$
= $2.0 \times 10^{6} m s^{- 1}$

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Similar questions

Q. In a chamber , a uniform magnetic field of

6.5 G (1 G = 10^{- 4} T)

is maintained. An electron is shot into the field with a speed of

4.8 \times 10^{6} m s^{- 1}

normal to the field.Explain why the path of the electron is a circle.Determine the radius of the circular orbit.(

e = 1.6 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} k g

)

Q. In a chamber, a uniform magnetic field of 6.5 G (1 G = 10–4 T) ismaintained. An electron is shot into the field with a speed of4.8 × 106 m s–1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.(e = 1.5 × 10–19 C, me = 9.1×10–31 kg)

An electron of mass $0.90 \times 10^{- 30} k g$ under the action of a magnetic field moves in a circle of 2.0 $c m$ radius at a speed of $3.0 \times 10^{6} m s^{- 1}$ . If a proton of mass $1.8 \times 10^{- 27} k g$ has to move in a circle of the same radius in the same magnetic field, then its speed will be $m s^{- 1}$ .

In a chamber, a uniform magnetic field of

6.5

G

(1 G = 10^{- 4} T)

is maintained. An electron is shot into the field with a speed of

4.8 \times 10^{6} m s^{- 1}

normal to the field. Explain why the path of the electron is a circle?

Determine the radius of the circular orbit.

(e = 1.5 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} k g)

Also obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Q. An electron of mass

0.90 \times 10^{- 30}

kg under the action of a magnetic field moves in a circular of 2.0 cm radius at a speed of

3.0 \times 10^{6} m s^{- 1}

. If a proton of mass

1.8 \times 10^{27}

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

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An electron having energy 10eV is circulating in a path of radius 0.1m having a magnetic field of 10−4T. The speed of the electron will be :

The correct option is A 2.0×(106)ms−1Magnetic force provides necessary centripetal force,qVB=mv2rqVB=2KErV=2KEqBr=2×1010−4×0.1=2.0×106ms−1

An electron having energy $10 e V$ is circulating in a path of radius $0.1 m$ having a magnetic field of $10^{- 4} T$ . The speed of the electron will be :

The correct option is A $2.0 \times (10^{6}) m s^{- 1}$
Magnetic force provides necessary centripetal force,
$q V B = \frac{m v^{2}}{r}$
$q V B = \frac{2 K E}{r}$
$V = \frac{2 K E}{q B r}$
= $\frac{2 \times 10}{10^{- 4} \times 0.1}$
= $2.0 \times 10^{6} m s^{- 1}$