Question

A charge of $1\text{C}$ is placed at one end of a non-conducting rod of length $0.6\text{m}$. The rod is rotated in a vertical plane about a horizontal axis passing through the other end of the rod with angular frequency ${10}^4\pi \text{rad/s}$. The magnetic field at a point on the axis of rotation at a distance $0.8\text{m}$ from centre of the circular path will be:

• A. $1.13\times {10}^{-3}\text{T}$
• B. $2.44\times {10}^{-3}\text{T}$
• C. $1.75\times {10}^{-3}\text{T}$
• D. $3.25\times {10}^{-3}\text{T}$

Answer 1

The correct option is A $1.13\times {10}^{-3}\text{T}$


The rotation of the rod will lead to an equivalent current in the circular path, because the charge $q$ will pass through any point on the circle after a time interval of one time period $\left(T\right)$.

$\therefore i=\frac{q}{T}=qf$

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

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

The magnetic field at point $\text{P}$ due to equivalent current carrying circular loop at axis will be.

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

Here, $R=0.96\text{m};x=0.8\text{m};i=5\times {10}^3\text{A}$

$B_P=\frac{4\pi \times {10}^{-7}\times (5\times {10}^3)\times (0.6{)}^2}{2\left[\right(0.6{)}^2+(0.8{)}^2{]}^{3/2}}$

$B_P=\frac{4\pi \times {10}^{-7}\times 5\times {10}^3\times 36\times {10}^{-2}}{2\times (1{)}^{3/2}}$

$\therefore B_p=1.13\times {10}^{-3}\text{T}$

Why this Question ? Tip: The pivoted end of the rod serves as the centre of the circular path in which charge $q$ is rotating.