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Magnetics question

2023 · 15 Apr · Shift 1 · Q63
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Magnetics question

2023 · 15 Apr · Shift 1 · Q63

JEE MainPhysicsMagneticsNumerical+4 / −1
An electron in a hydrogen atom revolves around its nucleus with a speed of 6.76×106 ms−16.76 \times 10^6 \mathrm{~ms}^{-1}6.76×106 ms−1 in an orbit of radius 0.52 A∘0.52 \mathrm{~A}^{\circ}0.52 A∘. The magnetic field produced at the nucleus of the hydrogen atom is ‾\underline{\hspace{2cm}}​ T.
Numerical answer
View written solutionFree

Correct answer: 40

  1. Model the revolving electron as a current loop

An electron moving in a circular orbit constitutes a current.

If the electron has speed vvv in an orbit of radius rrr, then the time period is

T=2πrvT = \frac{2\pi r}{v}T=v2πr​

So the current is

I=eT=ev2πrI = \frac{e}{T} = \frac{ev}{2\pi r}I=Te​=2πrev​
  1. Magnetic field at the center of a circular current loop

For a circular loop of radius rrr, the magnetic field at the center is

B=μ0I2rB = \frac{\mu_0 I}{2r}B=2rμ0​I​

Substituting III:

B=μ02r⋅ev2πr=μ0ev4πr2B = \frac{\mu_0}{2r}\cdot \frac{ev}{2\pi r} = \frac{\mu_0 ev}{4\pi r^2}B=2rμ0​​⋅2πrev​=4πr2μ0​ev​

Since

μ04π=10−7\frac{\mu_0}{4\pi} = 10^{-7}4πμ0​​=10−7

we get

B=10−7⋅evr2B = 10^{-7}\cdot \frac{ev}{r^2}B=10−7⋅r2ev​
  1. Substitute the given values

Given:

  • e=1.6×10−19 Ce = 1.6\times 10^{-19}\,\text{C}e=1.6×10−19C
  • v=6.76×106 m s−1v = 6.76\times 10^6\,\text{m s}^{-1}v=6.76×106m s−1
  • r=0.52 A˚=0.52×10−10 mr = 0.52\,\text{\AA} = 0.52\times 10^{-10}\,\text{m}r=0.52A˚=0.52×10−10m

Now,

r2=(0.52×10−10)2=0.2704×10−20=2.704×10−21r^2 = (0.52\times 10^{-10})^2 = 0.2704\times 10^{-20} = 2.704\times 10^{-21}r2=(0.52×10−10)2=0.2704×10−20=2.704×10−21

Also,

ev=(1.6×10−19)(6.76×106)=10.816×10−13=1.0816×10−12ev = (1.6\times 10^{-19})(6.76\times 10^6) = 10.816\times 10^{-13} = 1.0816\times 10^{-12}ev=(1.6×10−19)(6.76×106)=10.816×10−13=1.0816×10−12

Therefore,

B=10−7⋅1.0816×10−122.704×10−21B = 10^{-7}\cdot \frac{1.0816\times 10^{-12}}{2.704\times 10^{-21}}B=10−7⋅2.704×10−211.0816×10−12​ B=10−7⋅(1.08162.704×109)B = 10^{-7}\cdot \left(\frac{1.0816}{2.704}\times 10^9\right)B=10−7⋅(2.7041.0816​×109) 1.08162.704=0.4\frac{1.0816}{2.704} = 0.42.7041.0816​=0.4

So,

B=10−7⋅0.4×109=0.4×102=40 TB = 10^{-7}\cdot 0.4\times 10^9 = 0.4\times 10^2 = 40\,\text{T}B=10−7⋅0.4×109=0.4×102=40T
  1. Final answer

The magnetic field produced at the nucleus is

40 T\boxed{40\,\text{T}}40T​

So the required integer is

40\boxed{40}40​
  1. Comparison with stored answer

Stored correct answer = 404040

Our derived answer also is 404040, so they agree.

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