Skip to main content

PDoubt No 000001 - JEE


Question : An electron gun is placed inside a long solenoid of radius $R$ on its axis. The solenoid has $n \text{ turns/length}$ and carries a current $I$. The electron gun shoots an electron along the radius of the solenoid with speed $v$. If the electron does not hit the surface of the solenoid, maximum possible value of $v$ is (all symbols have their standard meaning) :

(a) $\dfrac{2e\mu_0nIR}{m}$
(b) $\dfrac{e\mu_0nIR}{4m}$
(c) $\dfrac{e\mu_0nIR}{2m}$
(d) $\dfrac{e\mu_0nIR}{m}$

Solution : Top view of solenoid

As we know maximum possible radius of electron $=\dfrac{R}{2}$
$\implies\dfrac{R}{2}=\dfrac{mv}{qB}=\dfrac{mv_{max}}{e(\mu_0nI)}$
$\therefore v_{max}=\dfrac{R}{2}\dfrac{e\mu_0nI}{m}$
$\fbox{$\therefore v_{max}=\dfrac{e\mu_0nIR}{2m}$}$
Labels:

Comments

Post a Comment

Popular posts from this blog

JEE MAIN 2021 March Attempt Mathematics Q1

Question 1: The middle term in the expansion of $\left(x^2+\dfrac{1}{x^2}+2\right)^{n}$ is (a) $\dfrac{n!}{\left(\dfrac{n}{2}!\right)^2}$ (b) $\dfrac{(2n)!}{\left(\dfrac{n}{2}!\right)^2}$ (c) $\dfrac{(2n)!}{\left(n!\right)^2}$ (d) $\dfrac{1.3.5\ldots(2n+1)}{n!}2^{n}$ Solution :
Post a Comment
Read more

Daily Question 108

Daily Question 108 $\textbf{Q108.}$ If $f(x)$ is a periodic function having period $7$ and $g(x)$ is periodic function having period $11$ then the period of $D(x)=\begin{vmatrix} f(x) & f\left(\dfrac{x}{3}\right)\\ g(x) & g\left(\dfrac{x}{5}\right)\\ \end{vmatrix}$ is (A) $231$ (B) $385$ (C) $1155$ (D) $77$ $\textbf{Ans.} (C)$ $\textbf{Sol.}$ Given $D(x)=\begin{vmatrix} f(x) & f\left(\dfrac{x}{3}\right)\\ g(x) & g\left(\dfrac{x}{5}\right)\\ \end{vmatrix}$ $\implies D(x)=f(x)g\left(\dfrac{x}{5}\right)-g(x)f\left(\dfrac{x}{3}\right)$ Period of $f(x)g\left(\dfrac{x}{5}\right)$ is $7\times 55 = 385$ Period of $g(x)f\left(\dfrac{x}{3}\right)$ is $11\times 21 = 231$ Hence the period of $D(x) = \text{LCM}$ of $(385, 231)=1155$ Compiled Daily Questions 1-100
Post a Comment
Read more