A copper wire with a radius of 0.1 mm and a resistance of 1 k is connected across a 20 V power supply. Determine the current density in the wire.

Correct answer: B. 6.37 × 10^5 A/m^2 Given, radius of the wire, r = 0.1 mm = 10^ -4 m Resistance of the wire, R = 1 k = 10^3 Voltage across the wire, V = 20 V Current in the wire is given by Ohm's law: I = (V)/(R) = (20)/(10^3) = 2 × 10^ -2 A Cross-sectional area of the wire: A = π r^2 = π (10^ -4 )^2 = π × 10^ -8 m^2 Current density J is the current per unit area: J = (I)/(A) = 2 × 10^ -2 π × 10^ -8 = (2)/(π) × 10^6 A/m^2 J (2)/(3.1416) × 10^6 0.6366 × 10^6 A/m^2 = 6.37 × 10^5 A/m^2

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Electric Current in Conductors

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Electric Current in Conductors

A copper wire with a radius of 0.1 mm and a resistance of 1 k is...

A copper wire with a radius of

0.1 \text{ mm}

and a resistance of

1 \text{ k}\Omega

is connected across a

20 \text{ V}

power supply. Determine the current density in the wire.

Options

A $2.00 \times 10^6 \text{ A/m}^2$
B $6.37 \times 10^5 \text{ A/m}^2$
C $6.37 \times 10^7 \text{ A/m}^2$
D $3.18 \times 10^5 \text{ A/m}^2$

Correct answer

B. $6.37 \times 10^5 \text{ A/m}^2$

Step-by-step solution

Given, radius of the wire,

r = 0.1 \text{ mm} = 10^{-4} \text{ m}

Resistance of the wire,

R = 1 \text{ k}\Omega = 10^3 \Omega

Voltage across the wire,

V = 20 \text{ V}

Current in the wire is given by Ohm's law:

I = \dfrac{V}{R} = \dfrac{20}{10^3} = 2 \times 10^{-2} \text{ A}

Cross-sectional area of the wire:

A = \pi r^2 = \pi (10^{-4})^2 = \pi \times 10^{-8} \text{ m}^2

Current density

J

is the current per unit area:

J = \dfrac{I}{A} = \dfrac{2 \times 10^{-2}}{\pi \times 10^{-8}} = \dfrac{2}{\pi} \times 10^6 \text{ A/m}^2

J \approx \dfrac{2}{3.1416} \times 10^6 \approx 0.6366 \times 10^6 \text{ A/m}^2 = 6.37 \times 10^5 \text{ A/m}^2

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