Question

A metallic conductor of irregular cross-section is as shown in the figure. A constant potential difference is applied across the ends (1) and (2). Then :

• A. the current at the cross-section $P$ equals the current at the cross-section $Q$
• B. the electric field intensity at $P$ is less than that at $Q$.
• C. the rate of heat generated per unit time at $Q$ is greater than that at $P$
• D. the number of electrons crossing per unit area of cross-section at $P$ is less than that at $Q$.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are (A), (B), (C), and (D).

Answer 2

For a steady current in a conductor, the current is the same through all cross-sections due to the conservation of charge. Therefore, the current at $P$ equals the current at $Q$.

Current density $J = I/A$. Since the cross-sectional area at $P$ ($A_P$) is larger than at $Q$ ($A_Q$), the current density at $P$ ($J_P$) is less than at $Q$ ($J_Q$).

The electric field intensity $E$ is related to current density by $J = \sigma E$, where $\sigma$ is the conductivity. Since $J_P < J_Q$ and $\sigma$ is constant for the conductor, $E_P < E_Q$.

The rate of heat generation per unit time (power dissipation) is given by $P = I^2R$. Resistance $R = \rho L/A$, where $\rho$ is resistivity. So, power dissipated per unit length is $dP/dL = I^2\rho/A$. Since $A_Q < A_P$, the rate of heat generated per unit length at $Q$ is greater than at $P$. Alternatively, power dissipated per unit volume is $P_v = J \cdot E = J^2/\sigma = (I/A)^2/\sigma$. Since $A_Q < A_P$, $P_v(Q) > P_v(P)$.

The number of electrons crossing per unit area of cross-section per unit time is the definition of current density. As established, $J_P < J_Q$. Thus, the number of electrons crossing per unit area at $P$ is less than that at $Q$.