Current Density: MCQs

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Current Density MCQ with answers and detailed explanations are given in this article. Current Density refers to the amount of electric charge that flows across a unit cross-section area in a unit of time. It is a vector quantity. The electric current density, denoted by J, for a given conducting material is given by the formula –

J = ${q/t \over A} ={ I \over A}$

where q refers to the charges flowing,
t is the time
A is the cross-sectional area
and I is the current

MCQ on Current Density and its Formula

Ques. The SI unit of current density is –

J/m²
I/m²
N/m²
A/m²

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Ans. d) A/m²

Explanation: As current density = I/A

The SI units for current and area are A and m². Therefore, the SI unit of current density is A/m².

Ques. A soft graphite brush has a contact area of 5 cm² and if the current density of the material is 9 A/cm², how much current will it be able to carry?

60 A
45 A
32 A
55 A

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Ans. b) 45 A

Explanation: As the current density, J = I/A

Given: J = 9 A/cm²

A = 5 cm²

9 = I/5

I = 45 A

Ques. An electric bulb filament carries a 0.75 A current in 9 minutes. Find the electric charge flowing through the current.

405 C
240 C
225 C
290 C

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Ans. a) 405 C

Explanation: Current (I) = 0.75 A

Time (t) = 9 minutes

Time can be converted from minutes into seconds as follows –

t = 9 x 60

= 540 sec

Q = I x t

Q = 0.75 x 540

Q = 405 C

Ques. What is the current density when 40 amperes current flows through the battery in a given area of 10 m²?

4 A/m²
7 A/m²
9 A/m²
2 A/m²

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Ans. a) 4 A/m²

Explanation: I = 40 A,

Area = 10 m²

The current density formula is J = I / A

= 40/10

J = 4 A/m².

Ques. The electric field E, conductivity σ of a conductor, and current density J are related as:

σ = E/J
σ = J/E
σ = EJ
σ = 1/JE

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Ans. b) σ = J/E

Explanation: Let I be the current that flows through a conductor when the V volt potential difference is applied. then, from Ohm’s law,
I = V/R

But R = $\rho$L/A

Hence, $I = { V \over \rho {L\over A}}\\ {I \over A} = { V \over \rho L} $

But I/A = J and V/L = E

Therefore, σ = J/E

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Ques. The electric field intensity E, specific resistance k, and current density J are related to each other as –

E = J/K
E = Jk
E = k/J
k = JE

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Ans. b) E = Jk

Explanation: We know that J = I/A and E = I/σA

Therefore, E = J/σ = kJ

Therefore, E = Jk

Ques. Calculate the current density when 30 amperes current flows through the battery of 5 m².

4 A/m²
3 A/m²
9 A/m²
6 A/m²

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Ans. d) 6 A/m²

Explanation: I = 30 A,

Area = 5 m²

The current density formula is J = I / A

= 30/5

J = 6 A/m².

Ques. Calculate the current density when 20 amperes current flows through the battery of 0.2 m².

50 A/m²
100 A/m²
30 A/m²
110 A/m²

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Ans. b) 100 A/m²

Explanation: I = 20 A,

Area = 0.2 m²

The current density formula is J = I / A

= 20/0.2

J = 100 A/m².

Ques. In copper, there are 10²⁸ electrons in a unit cubic meter, all of which contribute to a current of 2 A in the wire of copper of 1x 10^-6 m² cross-sectional area. What is the electric field in the wire? Given resistivity, ρ = 1.6 x 10^-8 Ωm.

3.2 x 10^-2 Vm^-1
2.2 x 10^-2 Vm^-1
1.5 x 10^-2 Vm^-1
4.6 x 10^-2 Vm^-1

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Ans. a) 3.2 x 10^-2 Vm^-1

Explanation: Given,

The density of electron, n = 10²⁸ m^-3
Current, I = 2A
Area, A = 1 x 10^-6 m²
Resistivity, ρ = 1.6 x 10^-8 Ωm

We have, Current density, J = I/A = 2/1 x 10^-6 = 2 x 10⁶ Am^-2

Also, J = σE

⇒ E = J/σ = ρJ

⇒ E = 1.6 x 10^-8 x 2 x 10⁶ = 3.2 x 10^-2 Vm^-1

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Ques. What is mobility?

Ease of carrier drift
Ease of movement
Ease of access to the junction
Ease of current flow

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Ans. a) Ease of carrier drift

Explanation: Mobility refers to the ease with which the carriers can drift. The higher the time of the collision, the greater the mobility is and hence the lighter the carrier, the greater its mobility.

Ques. The equation J_n= qnµ_nE (A/cm²) represents

Drift current density
Diffusion current density
Diffusion current
Drift current

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Ans. a) Drift current density

Explanation: In this equation, ‘q’ refers to the charge on the carrier, ‘n’ is the number of carriers, ‘µ_n’ is the mobility constant, and ‘E’ refers to the electric field intensity. These are the factors that comprise the drift current and therefore the given equation represents the drift current density.

Ques. What is the number of electrons which constitutes one Ampere?

6.25 × 10^-18
6.25 × 10¹⁸
2.25 × 10^-18
2.25 × 10¹⁸

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Ans. b) 6.25 × 10¹⁸

Explanation: Number of electrons, n = $\frac {Current (I) \times time (t)}{Charge \, of \, 1 \, electron (e)}$

$\frac {(1 \times 1)}{1.6 \times 10^{-19}}$

= 6.25 × 10¹⁸

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