Question

An electron is moving in a stable circular orbit of radius $0.1\text{ m}$ around a thin infinitely long positively charged straight wire. If the orbital velocity of the electron around the wire is $4 \times 10^7\text{ ms}^{-1}$, then the linear charge density of the wire is nearly:}

• A. $4.5 \times 10^{-7}\text{ Cm}^{-1}$
• B. $9 \times 10^{-7}\text{ Cm}^{-1}$
• C. $5 \times 10^{-7}\text{ Cm}^{-1}$
• D. $2.5 \times 10^{-7}\text{ Cm}^{-1}$

Hint:

For an electron moving in a circular orbit around an infinitely long charged wire, \[ eE=\frac{mv^2}{r} \] and \[ E=\frac{\lambda}{2\pi\varepsilon_0 r}. \] After substitution, the orbital radius cancels automatically, giving \[ \lambda=\frac{2\pi\varepsilon_0 mv^2}{e}. \] This is a frequently used shortcut in problems involving circular motion around a line charge.

Answer 1

The Correct Option is C

Concept: An infinitely long straight wire carrying a uniform linear charge density $\lambda$ produces an electric field around it. The magnitude of the electric field at a perpendicular distance $r$ from the wire is obtained using Gauss's Law and is given by \[ E=\frac{\lambda}{2\pi\varepsilon_0 r}. \] Since the wire is positively charged, the electric field is directed radially outward from the wire. An electron moving around the wire experiences an attractive electrostatic force because the electron is negatively charged. This attractive force continuously pulls the electron toward the wire and acts as the necessary centripetal force required for circular motion. Therefore, for a stable circular orbit, \[ \text{Electrostatic Force} = \text{Centripetal Force}. \] This force balance condition enables us to determine the unknown linear charge density of the wire.

Step 1: Write the expression for the electric field due to an infinitely long charged wire. According to Gauss's Law, \[ E=\frac{\lambda}{2\pi\varepsilon_0 r}. \] Here, \[ \lambda=\text{linear charge density}, \] \[ \varepsilon_0=\text{permittivity of free space}, \] and \[ r=\text{distance of the electron from the wire}. \] The electron experiences an electrostatic force \[ F_e=eE. \] Substituting the value of $E$, \[ F_e = e\left(\frac{\lambda}{2\pi\varepsilon_0 r}\right). \] Thus, \[ F_e = \frac{e\lambda}{2\pi\varepsilon_0 r}. \]

Step 2: Write the expression for the centripetal force. For a particle of mass $m_e$ moving in a circular orbit of radius $r$ with speed $v$, \[ F_c=\frac{m_e v^2}{r}. \] Since the orbit is stable, \[ F_e=F_c. \] Therefore, \[ \frac{e\lambda}{2\pi\varepsilon_0 r} = \frac{m_e v^2}{r}. \] Notice that the radius $r$ appears on both sides and cancels out completely. Hence, \[ \frac{e\lambda}{2\pi\varepsilon_0} = m_e v^2. \] This is an important result because it shows that the required charge density is independent of the orbital radius.

Step 3: Rearrange the equation to obtain $\lambda$. Multiplying both sides by \[ \frac{2\pi\varepsilon_0}{e}, \] we get \[ \lambda = \frac{2\pi\varepsilon_0 m_e v^2}{e}. \] Using \[ \frac{1}{4\pi\varepsilon_0} = 9\times10^9, \] we obtain \[ 2\pi\varepsilon_0 = \frac{1}{2\times9\times10^9}. \] Substituting this into the expression, \[ \lambda = \frac{m_e v^2} {2\left(\frac{1}{4\pi\varepsilon_0}\right)e}. \] This form is convenient for numerical calculations.

Step 4: Substitute all numerical values. Given, \[ m_e=9\times10^{-31}\text{ kg}, \] \[ v=4\times10^7\text{ ms}^{-1}, \] \[ e=1.6\times10^{-19}\text{ C}, \] and \[ \frac{1}{4\pi\varepsilon_0} = 9\times10^9. \] Substituting, \[ \lambda = \frac{(9\times10^{-31})(4\times10^7)^2} {2(9\times10^9)(1.6\times10^{-19})}. \] First calculate \[ (4\times10^7)^2 = 16\times10^{14}. \] Hence, \[ \lambda = \frac{(9\times10^{-31})(16\times10^{14})} {18\times10^9\times1.6\times10^{-19}}. \] Multiplying the numerator, \[ 9\times16=144. \] Therefore, \[ \lambda = \frac{144\times10^{-17}} {28.8\times10^{-10}}. \]

Step 5: Simplify the numerical value. \[ \frac{144}{28.8}=5. \] Also, \[ 10^{-17}\div10^{-10} = 10^{-7}. \] Thus, \[ \lambda = 5\times10^{-7}\text{ Cm}^{-1}. \] Final Conclusion: The linear charge density of the infinitely long positively charged wire is \[ \boxed{\lambda=5\times10^{-7}\text{ Cm}^{-1}}. \] Hence, the correct answer is \[ \boxed{\text{(C)}\;5\times10^{-7}\text{ Cm}^{-1}}. \]