Physics Q47 – JEE-MAIN 2026

Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the

2^{n d}

and

4^{t h}

orbits of hydrogen atom _______.

Enter Calculated Numerical Value:

Clue 1 of 3: Fundamental Principles

Recall the formulas for the radius and velocity of an electron in Bohr's model for the $n^{t h}$ orbit of a hydrogen atom.

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Step 1: Relate magnetic field to orbital parameters✦ Active

The magnetic field $B$ at the center of a circular orbit of radius $r$ due to an electron moving with velocity $v$ is given by $B = \frac{μ_{0} I}{2 r}$ . The current $I$ due to the electron's motion is $I = \frac{e v}{2 π r}$ . Substituting $I$ into the magnetic field formula, we get:

B = \frac{μ_{0}}{2 r} (\frac{e v}{2 π r}) = \frac{μ_{0} e v}{4 π r^{2}}

Step 2: Express velocity and radius in terms of principal quantum number

n

○ Expand

According to Bohr's model for a hydrogen atom ( $Z = 1$ ):

r_{n} = n^{2} a_{0} ⟹ r_{n} \propto n^{2}

v_{n} = \frac{v_{1}}{n} ⟹ v_{n} \propto \frac{1}{n}

💡 Teacher's Secret Hint

Remember that $a_{0}$ is the Bohr radius and $v_{1}$ is the velocity in the first orbit, both constants for a hydrogen atom.

Step 3: Determine the proportionality of

B_{n}

with

n

and calculate the ratio○ Expand

Substitute the proportionalities for $v_{n}$ and $r_{n}$ into the expression for $B_{n}$ :

B_{n} \propto \frac{v_{n}}{r_{n}^{2}} \propto \frac{1 / n}{(n^{2})^{2}} = \frac{1 / n}{n^{4}} = \frac{1}{n^{5}}

Now, calculate the ratio of the magnetic fields for the $2^{n d}$ and $4^{t h}$ orbits:

\frac{B_{2}}{B_{4}} = \frac{1 / 2^{5}}{1 / 4^{5}} = \frac{4^{5}}{2^{5}} = {(\frac{4}{2})}^{5} = 2^{5} = 32

💡 Teacher's Secret Hint

Ensure to correctly handle the powers of $n$ when deriving the proportionality for $B_{n}$ .

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Physics Question 47 – JEE-MAIN 2026

Clue 1 of 3: Fundamental Principles

🧠 Full Solution Path