Question

A charge of $3.14\times {10}^{-6}\text{C}$ is distributed uniformly over a circular ring of radius $20\text{cm}$. The ring rotates about its axis with angular velocity of $60\text{rad/s}$. Find the ratio of electric field to the magnetic field at a point on the axis located at a distance of $5\text{cm}$ from centre.

• A. $1.88\times {10}^{15}\text{m/s}$
• B. $2.50\times {10}^{15}\text{m/s}$
• C. $3.20\times {10}^{15}\text{m/s}$
• D. $4.50\times {10}^{15}\text{m/s}$

Answer 1

The correct option is A $1.88\times {10}^{15}\text{m/s}$

Given, $q=3.14\times {10}^{-6}\text{C};R=20\text{cm}=0.2\text{m}$ and $\omega =60\text{rad/s}$

Equivalent current in circular loop,

$i=qf=\frac{q\omega }{2\pi }=\frac{3.14\times {10}^{-6}\times 60}{2\pi }$

$i=30\times {10}^{-6}\text{A}$

The value of magnetic field at point $x=5\times {10}^{-2}\text{m}$ on axis is,

$B_{axis}=\frac{{\mu }_{0}i{R}^2}{2(R^2+x^2{)}^{3/2}}$

$B_a=\frac{4\pi \times {10}^{-7}\times 30\times {10}^{-6}\times (0.2{)}^2}{2{\left[\right(0.2{)}^2+\left(0.05{)}^2\right]}^{3/2}}$

$B_a=\frac{2\pi \times {10}^{-7}\times 30\times {10}^{-6}\times (0.2{)}^2}{{\left[\right(0.2{)}^2+\left(0.05{)}^2\right]}^{3/2}}$

Now, the value of electric field due to a ring distribution of charge is,

$E=\frac{kqx}{(R^2+x^2{)}^{3/2}}$

$E=\frac{9\times {10}^9\times 3.14\times {10}^{-6}\times \left(0.05\right)}{\left[\right(0.20{)}^2+(0.05{)}^2{]}^{3/2}}$

Thus,

$\frac{E}{B_{axis}}=\frac{(9\times {10}^9)\times (3.14\times {10}^{-6})\times \left(0.05\right)}{2\times 3.14\times {10}^{-7}\times 30\times {10}^{-6}\times (0.2{)}^2}$

$\frac{E}{B_{axis}}=\frac{9\times {10}^9\times 0.05}{{10}^{-7}\times 60\times 0.04}$

$\therefore \frac{E}{B_{axis}}=1.88\times {10}^{15}\text{m/s}$

Why this Question ? Note: The concept involved in this particular question is simple. Due to a ring, $E=\frac{kqx}{(R^2+x^2{)}^{3/2}}$ $B_{axis}=\frac{{\mu }_{0}i{R}^2}{2(R^2+x^2{)}^{3/2}}$ Caution: There are some questions that requires complicated calculations. These are often called "TRAP" questions. A smart student should always identify these trap questions and try to attempt it only after completing the entire paper, if the time permits