A long circular tube of length 10 ~m and radius 0.3 ~m carries a...

Take the circular tube as a long solenoid. The wires are closely wound. Magnetic field inside the solenoid is B=\mu_0 n i Here, $n=$ number of turns per unit length $\therefore n i=$ current per unit length In the given problem, $$\begin{aligned}& n i=\frac{I}{L}\\ \therefore \quad B & =\frac{\mu_0 I}{L} \end{aligned}$$Flux passing through the $circular coil is \phi=B S=$\left(\frac{\mu_0 I}{L}\right)\left(\pir^2\right) Induced emf,e=-\frac{d \phi}{d t}=-\left(\frac{\mu_0 \pi r^2}{L R}\right) \cdot \frac{d I}{d t}$Induced current, i=$\frac{e}{R}=-\left(\frac{\mu_0 \pi r^2}{L R}\right) \frac{d I}{d t}$Magnetic moment$i A=i \pi r^2$or$\quad M=-\left(\frac{\mu_0 \pi^2 r^4}{L R}\right) \cdot \frac{d I}{d t}$Given,$\quad I=I_0 \cos 300 t$\therefore \quad$\frac{d I}{d t}=-$300 I_0 \sin (300 t) Substituting$ in Eq. (i), we get$$\begin{aligned}M & =\left(\frac{300 \pi^2 r^4}{L R}\right) \mu_0 I_0 \sin 300 t \\ \therefore \quad N & =\frac{300 \pi^2 r^4}{L R} \end{aligned}$$$Substituting the values, we get$$\begin{aligned}N & =\frac{300(22 / 7)^2(0.1)^4}{(10)(0.005)}\\ & =5.926 \end{aligned}$$or \quad 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 \%.