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Magnetics question

2025 · 22 Jan · Shift 2 · Q56
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Magnetics question

2025 · 22 Jan · Shift 2 · Q56

JEE MainPhysicsMagneticsMCQ+4 / −1
An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr's quantization is applicable, then the radius of the electronic orbit in the first excited state is :
  1. A
    hπeB\sqrt{\frac{h}{\pi e B}}πeBh​​
  2. B
    4hπeB\sqrt{\frac{4 h}{\pi e B}}πeB4h​​
  3. C
    h2πeeB\sqrt{\frac{\mathrm{h}}{2 \pi e \mathrm{eB}}}2πeeBh​​
  4. D
    2hπeB\sqrt{\frac{2 h}{\pi e B}}πeB2h​​
View written solutionFree

Correct answer: A

  1. Magnetic force provides centripetal force

For an electron moving perpendicular to a uniform magnetic field BBB:

evB=mv2re v B = \frac{m v^2}{r}evB=rmv2​

So,

mvr=eBr2(1)mvr = eBr^2 \quad \text{(1)}mvr=eBr2(1)

  1. Apply Bohr quantization

According to Bohr's quantization of angular momentum,

mvr=nh2πmvr = n\frac{h}{2\pi}mvr=n2πh​

For the first excited state, n=2n=2n=2. Hence,

mvr=2h2π=hπ(2)mvr = 2\frac{h}{2\pi} = \frac{h}{\pi} \quad \text{(2)}mvr=22πh​=πh​(2)

  1. Equate (1) and (2)

From (1) and (2):

eBr2=hπeBr^2 = \frac{h}{\pi}eBr2=πh​

Therefore,

r2=hπeBr^2 = \frac{h}{\pi e B}r2=πeBh​

r=hπeBr = \sqrt{\frac{h}{\pi e B}}r=πeBh​​

  1. Match with options

This corresponds to:

hπeB\boxed{\sqrt{\frac{h}{\pi e B}}}πeBh​​​

So the correct option is A.

  1. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.

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