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Question

An electron is moving around a proton in an orbit of radius $$1\mathring{A}$$ and produces $$16Wb/{m}^{2}$$ of magnetic field at the centre, then find the angular velocity of electron:


A
20π×1016 rad/sec
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B
1017 rad/sec
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C
52π×1016 rad/sec
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D
54π×1016 rad/sec
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Solution

The correct option is B $$ {10}^{17}$$ rad/sec

$$B=16Wb/m^{2}$$
$$R=\mathring { A } $$
as $$B=\dfrac { { \mu  }_{ 0 }I }{ { 4 }\pi { R }^{ 2 } } \oint { dL } \\ B=\dfrac { { \mu  }_{ 0 }I }{ { 4 }\pi { R }^{ 2 } } \times 2\pi { R }=\dfrac { { \mu  }_{ 0 }I }{ 2{ R } } $$
as $$I=\dfrac { e }{ t } $$ or $$I=\dfrac { e\times \omega  }{ \pi  } $$ (as $$\dfrac { 2\pi  }{ \omega  } =t$$)
So, $$16=\dfrac { 4\pi \times { 10 }^{ -7 }\times \left( 1.6\times { 10 }^{ 19 } \right) \times \dfrac { \omega  }{ 2\pi  }  }{ 2\times { 10 }^{ -10 } } \\ \Rightarrow \omega =\dfrac { 16\times { 10 }^{ -10 } }{ 1.6\times { 10 }^{ -26 } } \\ \Rightarrow \omega ={ 10 }^{ 17 }$$rad/sec

Physics

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