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An electron revolves in a circle of radius 0.4`A^(0)` ith a speed of `10^6 ms^(-1)` in a hydrogen atom. The magnetic field produced at the centre of the orbit due to motion of the electron, in tesla, is `[mu_(0)=4pixx10^(-7)H//m]`

A

0.1

B

1

C

10

D

100

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The correct Answer is:
To find the magnetic field produced at the center of the orbit due to the motion of an electron in a hydrogen atom, we can follow these steps: ### Step-by-Step Solution 1. **Convert the radius from angstroms to meters**: \[ r = 0.4 \, \text{Å} = 0.4 \times 10^{-10} \, \text{m} = 4 \times 10^{-11} \, \text{m} \] 2. **Identify the speed of the electron**: \[ v = 10^6 \, \text{m/s} \] 3. **Calculate the time period (T) of the electron's revolution**: The formula for the time period \( T \) is given by: \[ T = \frac{2 \pi r}{v} \] Substituting the values: \[ T = \frac{2 \pi (4 \times 10^{-11})}{10^6} \] \[ T = \frac{8 \pi \times 10^{-11}}{10^6} = 8 \pi \times 10^{-17} \, \text{s} \] 4. **Calculate the current (I) due to the electron's motion**: The current \( I \) can be defined as the charge passing through a point per unit time. The charge of an electron \( e = 1.6 \times 10^{-19} \, \text{C} \). \[ I = \frac{e}{T} = \frac{1.6 \times 10^{-19}}{8 \pi \times 10^{-17}} \] 5. **Calculate the magnetic field (B) at the center of the orbit**: The formula for the magnetic field at the center of a circular loop due to current is given by: \[ B = \frac{\mu_0 I}{2r} \] Where \( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \). Substituting the values: \[ B = \frac{4\pi \times 10^{-7} \cdot \left(\frac{1.6 \times 10^{-19}}{8 \pi \times 10^{-17}}\right)}{2 \cdot (4 \times 10^{-11})} \] Simplifying this: \[ B = \frac{4\pi \times 10^{-7} \cdot 1.6 \times 10^{-19}}{16 \pi \times 10^{-11} \cdot 10^{-17}} \] \[ B = \frac{4 \cdot 1.6 \times 10^{-7 -19 + 17}}{16 \cdot 10^{-11}} = \frac{6.4 \times 10^{-9}}{16 \times 10^{-11}} = \frac{6.4}{16} \times 10^{2} = 0.4 \times 10^{2} = 10 \, \text{T} \] ### Final Answer The magnetic field produced at the center of the orbit due to the motion of the electron is: \[ B = 10 \, \text{T} \]
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AAKASH SERIES-MOVING CHARGES AND MAGNETISM-EXERCISE-II
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