A non-conducting thin disc of radius R rotate | vedclass

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

Question

(A) Consider a thin ring of thickness $dr$ at a distance $r$ from the center of the disc.The magnetic field $dB$ at the center due to this ring is giv

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(A) Consider a thin ring of thickness $dr$ at a distance $r$ from the center of the disc.The magnetic field $dB$ at the center due to this ring is given by $dB = \frac{\mu_0 dI}{2r}$.The current $dI$ due to the rotating ring is $dI = \frac{dQ}{T} = \frac{dQ}{2\pi} \omega = \frac{\sigma(r) (2\pi r dr)}{2\pi} \omega = \sigma(r) \omega r dr$.Substituting $\sigma(r) = \sigma_0 \left[1 + (\frac{r}{R})^\beta \right]$,we get $dI = \sigma_0 \omega \left[1 + (\frac{r}{R})^\beta \right] r dr$.Integrating to find the total magnetic field $B$ at the center:$B = \int_0^R \frac{\mu_0}{2r} \sigma_0 \omega \left[1 + (\frac{r}{R})^\beta \right] r dr = \frac{\mu_0 \sigma_0 \omega}{2} \int_0^R \left[1 + (\frac{r}{R})^\beta \right] dr$.$B = \frac{\mu_0 \sigma_0 \omega}{2} \left[ r + \frac{r^{\beta+1}}{R^\beta (\beta+1)} \right]_0^R = \frac{\mu_0 \sigma_0 \omega}{2} \left[ R + \frac{R}{\beta+1} \right] = \frac{\mu_0 \sigma_0 \omega R}{2} \left( \frac{\beta+2}{\beta+1} \right)$.Given $B = \frac{9}{10} \mu_0 \sigma_0 \omega R$,we equate:$\frac{1}{2} \left( \frac{\beta+2}{\beta+1} \right) = \frac{9}{10} \Rightarrow \frac{\beta+2}{\beta+1} = \frac{9}{5}$.$5\beta + 10 = 9\beta + 9 \Rightarrow 4\beta = 1 \Rightarrow \beta = \frac{1}{4}$.

Similar Questions

$PQRS$ is a square loop made of a uniform conducting wire. The current enters the loop at $P$ and leaves at $S$. Then the magnetic field will be

$A$ portion of a conductive wire is bent in the form of a semicircle of radius $r$ as shown in the figure. At the centre $O$ of the semicircle,the magnetic induction will be:

In the hydrogen atom,the electron is making $6.6 \times 10^{15} \, r.p.s.$ If the radius of the orbit is $0.53 \times 10^{-10} \, m,$ then the magnetic field produced at the centre of the orbit is (in $Tesla$):

The magnetic field due to a current-carrying circular coil on its axis at a distance of $\sqrt{2} \,d$ from the centre of the coil is $B$. If $d$ is the diameter of the coil,then the magnetic field at the centre of the coil is: (in $B$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module