An electron moves in a circular orbit with a uniform speed v. It produces a magnetic field B at the centre of the circle. The radius of the circle is proportional to:

Question

At which temperature will the r.m.s. velocity of a hydrogen molecule be equal to that of an oxygen molecule at \( 47^\circ \text{C} \)?

Answer 1

To determine the relationship between the radius of the circular orbit of the electron and the given variables, we start by understanding the physics involved when an electron moves in a circular orbit.

An electron moving in a circular orbit produces a magnetic field at the center due to its circular motion. The formula for the magnetic field at the center of a circular loop carrying current \(I\) is:

B = \frac{\mu_0 I}{2 r}

where:

B is the magnetic field.
\mu_0 is the permeability of free space.
I is the current.
r is the radius of the orbit.

For an electron moving with speed \(v\), the current \(I\) due to its motion is given by:

I = \frac{e \cdot v}{2 \pi \cdot r}

where \(e\) is the charge of the electron.

Substituting this expression for current \(I\) in the magnetic field formula, we have:

B = \frac{\mu_0 \cdot e \cdot v}{4 \pi \cdot r^2}

Rearranging this expression to solve for the radius \(r\), we get:

r^2 = \frac{\mu_0 \cdot e \cdot v}{4 \pi \cdot B}

Taking the square root of both sides, the radius \(r\) is given by:

r = \sqrt{\frac{\mu_0 \cdot e \cdot v}{4 \pi \cdot B}}

From this expression, you can see that:

r \propto \sqrt{\frac{v}{B}}

The radius of the circle is directly proportional to the square root of the velocity divided by the magnetic field. Therefore, the correct answer is:

\(\sqrt{\bigg(\frac{v}{B}\bigg)}\)

Answer 2