Problem 37 - A Complete Resource Book in Physics for JEE Main

Question

A non-conducting ring of radius \(R\) has charge \(Q\) distributed uniformly over it. If it rotates with an angular velocity \(\omega\), the equivalent current will be (A) Zero (B) \(Q \omega\) (C) \(Q \frac{\omega}{2 \pi}\) (D) \(Q \frac{\omega}{2 \pi R}\)

Step-by-Step Solution

Answer

The equivalent current for a non-conducting ring of radius R with a uniformly distributed charge Q rotating with an angular velocity ω is given by \(I = Q \frac{\omega}{2 \pi}\), which corresponds to option (C).

Step 1: Write the formula for current

We know that current (I) is defined as the rate of flow of charge (Q) with respect to time (t), which can be written as: \(I = \frac{Q}{t}\)

Step 2: Calculate time period of one rotation

Let's calculate the time period, denoted by T, for one complete rotation of the non-conducting ring. The time period can be expressed in terms of angular velocity (ω), using the following formula: \(T = \frac{2 \pi}{\omega}\)

Step 3: Substitute time period in the current formula

Now, we will substitute the time period T from step 2 into the formula for current (I) from step 1: \(I = \frac{Q}{\frac{2 \pi}{\omega}}\)

Step 4: Simplify the equation

By simplifying the equation from step 3, we can find the equivalent current (I): \(I = Q \frac{\omega}{2 \pi}\) Now, we can conclude that the equivalent current corresponds to option (C).