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Standard XII

Physics

Circular Kinematics

An electron i...

Question

An electron is projected with a velocity $3 \times 10^{10} m s^{- 1}$ at right angle to a uniform magnetic field of induction $0.2 T$ . The radius of the orbit that the electron revolves is
(Charge of $e = 1.6 \times 10^{- 19} C$ , $m_{e} = 9.1 \times 10^{- 31} k g$ )

0.68 m

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0.42 m

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0.24 m

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0.84 m

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Solution

The correct option is C $0.84 m$
$r = \frac{m v}{q B}$
= $\frac{9.1 \times 10^{- 31} \times 3 \times 10^{10}}{1.6 \times 10^{- 19} \times 0.2} = 0.84 m$

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Similar questions

Q. A cathode ray beam is bent into an arc of a circle of radius

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Q. In a chamber , a uniform magnetic field of

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Q. Suppose that the particle is an electron projected with velocity

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Suppose that the particle in Exercise in 1.33 is an electron projected with velocity v_x= 2.0 × 10⁶ m s⁻¹. If E between the plates separated by 0.5 cm is 9.1 × 10² N/C, where will the electron strike the upper plate? (| e | =1.6 × 10⁻¹⁹ C, m_e= 9.1 × 10⁻³¹ kg.)

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An electron is projected with a velocity 3×1010ms−1 at right angle to a uniform magnetic field of induction 0.2T. The radius of the orbit that the electron revolves is (Charge of e=1.6×10−19C, me=9.1×10−31kg )

The correct option is C 0.84mr=mvqB=9.1×10−31×3×10101.6×10−19×0.2=0.84m

An electron is projected with a velocity $3 \times 10^{10} m s^{- 1}$ at right angle to a uniform magnetic field of induction $0.2 T$ . The radius of the orbit that the electron revolves is
(Charge of $e = 1.6 \times 10^{- 19} C$ , $m_{e} = 9.1 \times 10^{- 31} k g$ )

The correct option is C $0.84 m$
$r = \frac{m v}{q B}$
= $\frac{9.1 \times 10^{- 31} \times 3 \times 10^{10}}{1.6 \times 10^{- 19} \times 0.2} = 0.84 m$