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A metallic wire of length $L$, uniform cross-sectional area $A$, and resistivity $\rho$ carries a steady current $I$ when a potential difference $V$ is applied across its ends. The free electron density in the wire is $n$, and the magnitude of the charge of an electron is $e$. The average drift velocity of the electrons is $v_d$. Which of the following statements is **incorrect** regarding the properties of this wire? A) The potential difference across the wire is given by $V = I \left(\frac{\rho L}{A}\right)$. B) The current flowing through the wire is given by $I = n A e v_d$. C) The magnitude of the uniform electric field inside the wire is $E = \frac{V}{L}$. D) The current density in the wire is given by $J = \frac{n e v_d}{A}$.
The correct answer is **D**. Let's analyze each option to determine the incorrect statement: **A) The potential difference across the wire is given by $V = I \left(\frac{\rho L}{A}\right)$.** According to Ohm's Law, the potential difference ($V$) across a conductor is related to the current ($I$) flowing through it by its resistance ($R$), i.e., $V = IR$. The resistance of a metallic wire of length $L$, uniform cross-sectional area $A$, and resistivity $\rho$ is given by the formula: $$R = \frac{\rho L}{A}$$ Substituting the expression for $R$ into Ohm's Law, we obtain: $$V = I \left(\frac{\rho L}{A}\right)$$ This statement is **correct**. **B) The current flowing through the wire is given by $I = n A e v_d$.** The relationship between the macroscopic current ($I$) and the microscopic drift velocity ($v_d$) of charge carriers (electrons) in a conductor is a fundamental equation in current electricity. Here, $n$ is the number density of free electrons, $A$ is the cross-sectional area of the conductor, and $e$ is the magnitude of the charge of an electron. Thus, $I = n A e v_d$. This statement is **correct**. **C) The magnitude of the uniform electric field inside the wire is $E = \frac{V}{L}$.** When a potential difference $V$ is applied across a uniform conductor of length $L$, a uniform electric field $E$ is established within the conductor. The magnitude of this electric field is given by the potential gradient: $$E = \frac{V}{L}$$ This electric field is responsible for accelerating the free electrons and causing the current flow. This statement is **correct**. **D) The current density in the wire is given by $J = \frac{n e v_d}{A}$.** Current density ($J$) is defined as the current per unit cross-sectional area. Its magnitude is given by: $$J = \frac{I}{A}$$ From option B, we know that $I = n A e v_d$. Substituting this expression for $I$ into the definition of current density yields: $$J = \frac{n A e v_d}{A}$$ $$J = n e v_d$$ The statement given in option D is $J = \frac{n e v_d}{A}$. This expression incorrectly includes the cross-sectional area $A$ in the denominator, whereas the correct expression for current density is $J = n e v_d$. Therefore, this statement is **incorrect**. Thus, the incorrect statement among the given options is D.
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