JEE Main 2022 · 24 Jun · Shift 2 · Physics Q46

← JEE Main 2022 · 24 Jun · Shift 2

PhysicsSingle correct+4 / -1Question 46 of 90

A long cylindrical volume contains a uniformly distributed charge of density $\rho$ . The radius of cylindrical volume is $R$ . A charge particle ( $q$ ) revolves around the cylinder in a circular path. The kinetic energy of the particle is :

\dfrac{\rho qR^2}{4\varepsilon_0}

\dfrac{\rho qR^2}{2\varepsilon_0}

\dfrac{q\rho}{4\varepsilon_0R^2}

\dfrac{4\varepsilon_0R^2}{q\rho}

Answer and worked solution

Official answer: A

Field outside the cylinder (Gauss's law, length $l$ ): $E(2\pi rl)=\dfrac{\rho\pi R^2l}{\varepsilon_0}$ , so $E=\dfrac{\rho R^2}{2\varepsilon_0r}$ .

Circular motion. $\dfrac{mv^2}r=qE=\dfrac{q\rho R^2}{2\varepsilon_0r}$ , so $mv^2=\dfrac{q\rho R^2}{2\varepsilon_0}$ .

Kinetic energy. $\frac12mv^2=\dfrac{\rho qR^2}{4\varepsilon_0}$ , independent of the orbit radius.

(A)

Topics:Electric flux and Gauss's law Circular motion dynamics

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JEE Main 2022 · 24 Jun · Shift 2, Physics question 46