A thin charged rod is bent into the shape of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The magnetic field on the axis of a current-carrying circular loop at a distance $d$ from its center is given by:$B = \frac{\mu_{0} I R^{2}}{2(R^{

Vedclass · Accepted Answer

$m=3, n=3$

Vedclass · Answer

(D) The magnetic field on the axis of a current-carrying circular loop at a distance $d$ from its center is given by:$B = \frac{\mu_{0} I R^{2}}{2(R^{2} + d^{2})^{3/2}}$Since the rod is rotated with a time period $T$, the equivalent current $I$ is:$I = \frac{q}{T} = \frac{\lambda (2 \pi R)}{T}$Substituting $I$ into the magnetic field formula:$B = \frac{\mu_{0} (2 \pi R \lambda / T) R^{2}}{2(R^{2} + d^{2})^{3/2}}$For $d >> R$, we can approximate $(R^{2} + d^{2})^{3/2} \approx (d^{2})^{3/2} = d^{3}$.Thus, the magnetic field becomes:$B \approx \frac{\mu_{0} (2 \pi R \lambda) R^{2}}{2 T d^{3}} = \frac{\mu_{0} \pi \lambda}{T} \frac{R^{3}}{d^{3}}$Comparing this with the given form $B \propto \frac{R^{m}}{d^{n}}$, we get $m = 3$ and $n = 3$.

Similar Questions

Two similar coils, which are separated by $2 \ m$, have a radius of $1 \ m$ and $80$ turns, and share a common axis. Calculate the strength of the magnetic field in microtesla at a point midway between them on their common axis when the current is $0.2 \ A$.

If an electron revolves around a nucleus in a circular orbit of radius $R$ with frequency $n$,then the magnetic field produced at the centre of the nucleus will be

Two concentric circular coils of $10$ turns each are situated in the same plane. Their radii are $20 \, cm$ and $40 \, cm$ and they carry respectively $0.2 \, A$ and $0.3 \, A$ current in opposite directions. The magnetic field at the centre is $(\mu_0 = 4 \pi \times 10^{-7} \, T \cdot m/A)$

$A$ dielectric circular disc of radius $R$ carries a uniform surface charge density $\sigma$. If it rotates about its axis with angular velocity $\omega$, the magnetic field at the center of the disc is:

If $B_1$ is the magnetic field induction at a point on the axis of a circular coil of radius $R$ situated at a distance $R \sqrt{3}$ and $B_2$ is the magnetic field at the centre of the coil,then the ratio of $\frac{B_1}{B_2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module