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Q.An electron is moving around an infinite linear charge in circular path of diameter 0.30 m. If linear charge density is 10^-6 C/m, then calculate the speed of an electron. (m_e = 9.0 x 10^-31 kg, e = 1.6 x 10^-19 C)

Rajasthan RbseRajasthan Board Senior Secondary Examination 2020Subjective· 2mImportance★★★★★
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The electric field of the infinite line charge supplies the centripetal force for the electron's circular motion; solving for v shows the radius cancels out, giving a speed independent of the orbit size.

Given: diameter =0.30 m⇒=0.30\,\text{m} \Rightarrow radius r=0.15 mr=0.15\,\text{m}; linear charge density λ=10−6 C/m\lambda=10^{-6}\,\text{C/m}; me=9.0×10−31 kgm_e=9.0\times10^{-31}\,\text{kg}; e=1.6×10−19 Ce=1.6\times10^{-19}\,\text{C}.

The electric field at perpendicular distance rr from an infinite line charge is:

E=λ2πε0rE=\dfrac{\lambda}{2\pi\varepsilon_0 r}

This field exerts an attractive electrostatic force on the electron (directed radially inward, towards the line charge), which provides the centripetal force needed for circular motion:

eE=mev2reE=\dfrac{m_e v^2}{r}

Substituting EE:

e⋅λ2πε0r=mev2re\cdot\dfrac{\lambda}{2\pi\varepsilon_0 r}=\dfrac{m_e v^2}{r}

The radius rr cancels from both sides:

v2=eλ2πε0mev^2=\dfrac{e\lambda}{2\pi\varepsilon_0 m_e}

Substituting values (ε0=8.85×10−12 F/m\varepsilon_0=8.85\times10^{-12}\,\text{F/m}):

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