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
B
C
D

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/secPhysics

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