Question

A charge of 1 $\text{coulomb}$ is placed at one end of a non-conducting rod of length $0.6\text{m}$. Half of the charge is removed from one end and placed on the other end. The rod is rotated in a vertical plane about a horizontal axis passing through the mid-point of the rod with angular frequency ${10}^4\pi \text{rad/s}$. The magnetic field at a point on the axis at a distance of $0.4\text{m}$ from centre of rod will be :

• A. $1.3\times {10}^{-3}\text{T}$
• B. $2.26\times {10}^{-3}\text{T}$
• C. $2.75\times {10}^{-3}\text{T}$
• D. $3.8\times {10}^{-3}\text{T}$

Answer 1

The correct option is B $2.26\times {10}^{-3}\text{T}$

If half of the charge is placed at the other end, then equivalent current in the circular path will develop due to rotation of both the charges placed at the two ends. So the radius of circular loop is,

$R=\frac{L}{2}=\frac{0.6}{2}=0.3\text{m}$

The distance of point $\text{P}$ from centre of loop is $x=0.4\text{m}$.


The equivalent current in circular loop will be

$i=\frac{\left(q/2\right)}{T}+\frac{\left(q/2\right)}{T}=\frac{q}{2}f+\frac{q}{2}f$

$i=qf=q\frac{\omega }{2\pi }=\frac{\left(1\right)\left({10}^4\pi \right)}{2\pi }$

$i=5\times {10}^3\text{A}$

Using the relation for magnetic field at a point on the axis of circular loop;

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

$B=\frac{(4\pi \times {10}^{-7})\times (5\times {10}^3)\times (0.3{)}^2}{2\left[\right(0.3{)}^2+\left(0.4{)}^2{]}^{3/2}\right]}$

$B=\frac{180\pi \times {10}^{-6}}{2{\left[0.5\right]}^3}=\frac{180\pi \times {10}^{-3}}{250}$

$B=2.26\times {10}^{-3}\text{T}$

Hence, option $\left(b\right)$ is correct.

Why this question ? Caution: The rod is non-conducting, hence the charge placed at respective ends of rod will remain at that position without getting dislocated unlike in the case of conducting rod.