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Standard IX

Physics

Fundamental and Derived Units

An electron a...

Question

An electron at point A in figure has a speed $v_{o} = 1.41 \times 10^{6} m / s$ . Find the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path from A to B,

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Solution

At A, magnetic force should be towards right. Thus from Fleming's left hand rule magnetic field should be inwards.
The electron experiences centripetal force under the action of magnetic field.
Hence $\frac{m v^{2}}{r} = q v B$
$⟹ B = \frac{m v}{q r}$
$= 1.6 \times 10^{- 4} T$

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Similar questions

Q. An electron at point A in the figure has a speed

V_{0}

of magnitude

1.41 \times 10^{6} m / s

. It enters into a uniform magnetic field and follows a semicircular path in it as shown.

The magnitude of the magnetic field that will cause the electron to follow the semicircular path from A to B is?

Q. An electron has a speed

\sqrt{2} \times 10^{6} m s^{- 1}

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as shown in Fig. Find the direction and magnitude of magnetic field so that electron reaches

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Q. An electron at point A in figure has a speed

v_{o} = 1.41 \times 10^{6} m / s

. Find the time required for the electron to move from A to B.

Answer the following questions:

(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?

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(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

Q. Refer to the figure, rectangle

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represents the cross section of a uniform magnetic field region of

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. An electron is projected at a speed of

v = 2.0 \times 10^{6} m / s

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An electron at point A in figure has a speed vo=1.41×106m/s. Find the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path from A to B,

At A, magnetic force should be towards right. Thus from Fleming's left hand rule magnetic field should be inwards.The electron experiences centripetal force under the action of magnetic field.Hence mv2r=qvB⟹B=mvqr=1.6×10−4T

An electron at point A in figure has a speed $v_{o} = 1.41 \times 10^{6} m / s$ . Find the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path from A to B,

At A, magnetic force should be towards right. Thus from Fleming's left hand rule magnetic field should be inwards.
The electron experiences centripetal force under the action of magnetic field.
Hence $\frac{m v^{2}}{r} = q v B$
$⟹ B = \frac{m v}{q r}$
$= 1.6 \times 10^{- 4} T$