JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Magnetic Effects of Current

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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 :

(A) $\frac{2 h}{π e B}$
(B) $\frac{4 h}{π eB}$
(C) $\frac{h}{2 π e B}$
(D) $\frac{h}{π eB}$ ✓ Correct

Answer: (D)

Solution

An electron moving perpendicular to a magnetic field B travels in a circle of radius r, with the magnetic force supplying the centripetal force:

e v B = \frac{m v ^{2}}{r} \Rightarrow m v = e B r

Bohr's quantization condition states that angular momentum is quantized:

m v r = \frac{nh}{2 π}

Substituting

m v = e B r

(e B r) r = \frac{nh}{2 π} \Rightarrow r^{2} = \frac{nh}{2 π e B} \Rightarrow r = \frac{nh}{2 π e B}

The first excited state corresponds to n = 2:

r = \frac{2 h}{2 π e B} = \frac{h}{π e B}

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Magnetic Effects of Current

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