Physics - Magnetism Question with Solution | TestHub

Take the circular tube as a long solenoid. The wires are closely wound. Magnetic field inside the solenoid is $$B={\mu }_{0}ni$$ Here, $n=$ number of turns per unit length $\therefore ni=$ current per unit length In the given problem, $$\begin{array}{rl}& ni=\frac{I}{L}\\ \therefore B& =\frac{{\mu }_{0}I}{L}\end{array}$$ Flux passing through the circular coil is $$\varphi =BS=\left(\frac{{\mu }_{0}I}{L}\right)\left(\pi r^2\right)$$ Induced emf, $e=-\frac{d\varphi }{dt}=-\left(\frac{{\mu }_0\pi r^2}{LR}\right)\cdot \frac{dI}{dt}$ Induced current, $$i=\frac{e}{R}=-\left(\frac{{\mu }_0\pi r^2}{LR}\right)\frac{dI}{dt}$$ Magnetic moment $iA=i\pi r^2$ or $M=-\left(\frac{{\mu }_0{\pi }^2{r}^4}{LR}\right)\cdot \frac{dI}{dt}$ Given, $I=I_0\cos 300t$ $$\therefore \frac{dI}{dt}=-300I_0\sin (300t)$$ Substituting in Eq. (i), we get $$\begin{array}{rl}M& =\left(\frac{300{\pi }^2{r}^4}{LR}\right){\mu }_0{I}_0\sin 300t\\ \therefore N& =\frac{300{\pi }^2{r}^4}{LR}\end{array}$$ Substituting the values, we get $$\begin{array}{rl}N& =\frac{300(22/7{)}^2(0.1{)}^4}{(10)(0.005)}\\ & =5.926\end{array}$$ or $N\simeq 6$ Analysis of Question Question is difficult to understand inside the examination hall. But 5 to $10\%$ question in IIT JEE are always difficult. Students should not panic. Because topper of IIT JEE scores approximately $80-90\%$.