A current of 5 A is passing through a non-linear magnesium wire of cross-section 0.04 m^2. At every

Q: Why do we use the dot product to find current?

Current is a scalar quantity that represents the total rate of charge flow. It is calculated as the flux of the current density vector $\vec{J}$ through an area $\vec{A}$, which mathematically is their dot product.

Q: What does the 'unit vector of area' mean?

The area vector is always perpendicular (normal) to the surface of the cross-section. Its direction indicates the orientation of the surface in space.

Q: Why can't we just use $V = IR$ here?

The macroscopic formula $V = IR$ requires the total length and voltage across the wire, which aren't provided. Instead, we use the microscopic form $E = \rho J$, which relates local properties at any specific point.

Q: Does the non-linear shape of the wire affect the calculation?

No, because the problem specifies the local conditions ('at every point') such as the cross-sectional area and the angle between $\vec{J}$ and $\vec{A}$. The microscopic Ohm's law applies point-by-point regardless of the overall shape.

Q: What is the direction of the electric field?

In an isotropic conductor, the electric field $\vec{E}$ is always parallel to the current density vector $\vec{J}$, as seen in the relation $\vec{E} = \rho \vec{J}$.

Question

A current of $5	ext{ A}$ is passing through a non-linear magnesium wire of cross-section $0.04	ext{ m}^2$. At every point, the direction of current density is at an angle of $60^\circ$ with the unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is
(Take, resistivity of magnesium, $ho = 44 	imes 10^{-8}\ \Omega	ext{-m}$)

Options:

(A) $11 	imes 10^{-2}	ext{ V/m}$

(B) $11 	imes 10^{-7}	ext{ V/m}$

(C) $11 	imes 10^{-5}	ext{ V/m}$

(D) $11 	imes 10^{-3}	ext{ V/m}$

Sigma Percentile · Accepted Answer

### The Flow of Charge: Macroscopic vs Microscopic

When we first learn about electricity, we usually talk about **current** ($I$) flowing through a wire, driven by a **voltage** ($V$) across a **resistance** ($R$). This is the macroscopic view of Ohm's Law: $V = IR$.

However, to truly understand what happens *inside* the material at any specific point, we need to zoom in. This brings us to the microscopic quantities: **Current Density** ($\vec{J}$) and **Electric Field** ($\vec{E}$).

### Current as a Flux

Current density $\vec{J}$ is a vector that tells us how much charge is flowing per unit time through a unit area perpendicular to the flow. But what if the area isn't perfectly perpendicular?

Imagine holding a net in a flowing river. If you hold it perfectly straight against the current, you catch the most water. If you tilt it, you catch less. The total water flowing through the net depends on the angle.

Mathematically, the total current $I$ is the **flux** of the current density vector through the cross-sectional area vector $\vec{A}$:
$$I = \vec{J} \cdot \vec{A} = JA \cos	heta$$

In our problem, the wire has a cross-section $A = 0.04 	ext{ m}^2$, and the current density is tilted at an angle $	heta = 60^\circ$ relative to the area vector. Given a total current $I = 5 	ext{ A}$, we can find the magnitude of the current density:
$$5 = J 	imes 0.04 	imes \cos(60^\circ)$$
$$5 = J 	imes 0.04 	imes \frac{1}{2}$$
$$J = \frac{5}{0.02} = 250 	ext{ A/m}^2$$

### The Microscopic Ohm's Law

Now that we know how intensely the charges are flowing ($\vec{J}$), we need to find the force driving them. The electric field $\vec{E}$ pushes the charges, and the material's **resistivity** ($ho$) resists this push.

The microscopic form of Ohm's Law beautifully connects these three local properties:
$$\vec{E} = ho \vec{J}$$

This equation tells us that the electric field required to maintain a certain current density is directly proportional to the material's resistivity.

### Final Calculation

We are given the resistivity of magnesium, $ho = 44 	imes 10^{-8} \Omega	ext{-m}$. Substituting our values into the microscopic Ohm's Law:
$$E = (44 	imes 10^{-8}) 	imes 250$$
$$E = 11000 	imes 10^{-8}$$
$$E = 11 	imes 10^{-5} 	ext{ V/m}$$

And there we have it! By understanding the vector nature of current density and the microscopic form of Ohm's Law, we've successfully navigated through the problem.

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Visualized Solution

$C u r r e n t an d A r e a V ec t or s$

$C u r r e n t an d C u r r e n t D e n s i t y$

$S u b s t i t u t in g V a l u es$

$C a l c u l a t in g C u r r e n t D e n s i t y$

$O h m^{'} s L a w in V ec t or F or m$

$C a l c u l a t in g E l ec t r i c F i e l d$

$C o n c l u s i o n$

$F oo d f or T h o ug h t$

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

The Flow of Charge

Current as a Flux

The Microscopic Ohm's Law

Final Calculation

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Comprehension Passage

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Current and Area Vectors

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Current and Current Density

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Current Density

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ohm′s Law in Vector Form

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Electric Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Food for Thought