Question

electron is rotating in a circular orbit with radius 5.2 into 10 raise to minus 11 metre and with linear Speed 2 into 10 raise to 6 metre per second in a hydrogen atom around the proton find the magnetic field produced at the centre of the earth​

Answer 1

Given:

Electron editing in a circular orbit with radius 5.2 × 10^(-11) m and linear velocity 2 × 10^(6) m/s in a hydrogen atom around a proton.

To find:

Magnetic Field produced at the centre of atom.

Calculation:

Applying Bio-Savart's Law:

$B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{i \: dl}{ {r}^{2} } \bigg \}$

$= > B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{( \frac{dq}{dt} ) \: dl}{ {r}^{2} } \bigg \}$

$= > B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{dq \times ( \frac{dl}{dt} )}{ {r}^{2} } \bigg \}$

$= > B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{dq \times v}{ {r}^{2} } \bigg \}$

Putting available values:

$= > B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{(1.6 \times {10}^{ - 19} )\times( 2 \times {10}^{6} )}{ {(5.2 \times {10}^{ - 11}) }^{2} } \bigg \}$

$= > B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{3.2 \times {10}^{ - 13} }{ 27.04\times {10}^{ - 22} } \bigg \}$

$= > B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ 0.132 \times {10}^{9} \bigg \}$

$= > B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ 13.2 \times {10}^{7} \bigg \}$

$= > B = {10}^{ - 7} \times \bigg \{ 13.2 \times {10}^{7} \bigg \}$

$= > B = 13.2 \: tesla$

So, final answer is:

$\boxed{ \bold{ \large{ \red{ B = 13.2 \: tesla}}}}$