Current Density MCQ Quiz - Objective Question with Answer for Current Density - Download Free PDF

In a metallic conductor with a steady current, the electric current remains constant throughout the conductor, even if the cross-sectional area varies. This is due to the conservation of charge.

However, the current density (J) and drift velocity (v_d) vary along the length of the conductor because they depend on the cross-sectional area (A) and resistivity, which can change in a non-uniform conductor.

For a conductor with non-uniform cross-section, the electric current I remains constant, but the current density J = I / A and drift velocity v_d = J / n e may change due to the variation in cross-sectional area (A).

∴ The physical quantity that remains constant is electric current.

The correct answer is: 4) Electric current.

Given:

Cross section area of the electron beam, A = 1.0 mm2 = 1.0 × 10 6 m2

Number of electrons passing per second, n = 6 × 1016

Charge of an electron, q = 1.6 × 10 19 C

Concept:

The current density, J, is given by the formula:

Formula Used:

J = I / A

Where 'I' is current and A is the cross sectional area.

The current, I, is given by:

I = n × q

Calculation:

⇒ I = 6 × 1016 × 1.6 × 10 19 C/s

⇒ I = 9.6 × 10 3 A

⇒ J = I / A

⇒ J = (9.6 × 10 3 A) / (1.0 × 10 6 m2)

⇒ J = 9.6 × 103 A/m2

Hence, the current density in the beam is 9.6 × 103 A/m2.

 Concept :

When two conductors are brought into contact, charge will redistribute between them until they reach the same electrical potential. For spheres, the surface charge density is related to the charge and the radius of the sphere. Before contact, both spheres have the same surface charge density. Let's denote this common surface charge density as σ.

For a sphere of radius R, the surface area A is given by:

A = 4πR²

The charge Q on each sphere can be expressed as:

Q = σ × A = 0.4πR²

Calculation:

Given the radii of the two spheres, the charge on each sphere before contact is:

For the sphere with radius R/2:

Q₁ = σ × 4π(R/2)² = σ × πR²

For the sphere with radius 2R:

Q₂ = σ × 4π(2R)² = σ × 16πR²

When the two spheres are brought into contact and then separated, the total charge will be redistributed between them. The total charge is:

Q_total = Q₁ + Q₂ = σ × πR² + σ × 16πR² = 17σ × πR²

Since the spheres are in contact, they will have the same potential. The potential (V) for a sphere is given by:

V = Q / R

Let Q₁' and Q₂' be the charges on the spheres with radii R/2 and 2R respectively after they have been separated. The potential of both spheres must be equal:

Q₁' / (R / 2) = Q₂' / (2R)

Simplifying, we get:

Q₁' × 2R = Q₂' × (R / 2) ⇒ Q₁' = Q₂' / 4

We also know that the total charge is conserved:

Q₁' + Q₂' = 17σ × πR²

Substituting Q₁' = Q₂' / 4 into this equation, we get:

(Q₂' / 4) + Q₂' = 17σ × πR² ⇒ 5Q₂' / 4 = 17σ × πR² ⇒ Q₂' = (68σ × πR²) / 5

And for the smaller sphere:

Q₁' = Q₂' / 4 = (68σ × πR²) / 5 × 1 / 4 = 17σ × πR² / 5

Now, we can find the new surface charge densities:

For the sphere with radius R/2:

σ₁' = Q₁' / (4π(R/2)²) = (17σ × πR² / 5) / (πR²) = 17σ / 5

Finally, the ratio of the new surface charge densities σ₁' : σ₂' is:

σ₁' / σ₂' = (17σ / 5) / (17σ / 20) = 20 / 5 = 4 : 1

So, the correct answer is Option 2: 4 : 1 

Calculation:

We know that

\(I=Ane{ v }_{ d }\)

\(\Rightarrow \dfrac { V }{ R } =Ane{ v }_{ d }\)

\(\Rightarrow \dfrac { V }{ \left( \rho \frac { l }{ A } \right) } =Ane{ v }_{ d }\)

\(\Rightarrow \dfrac { VA }{ \rho l } =Ane{ v }_{ d }\)

\(\Rightarrow \rho =\dfrac { V }{ lne{ v }_{ d } }\)

\(\Rightarrow \rho =\dfrac { 5 }{ 0.1\times \left( 8\times { 10 }^{ 28 } \right) \times \left( 1.6\times { 10 }^{ -19 } \right) \times \left( 2.5\times { 10 }^{ -4 } \right) } \)

\(\Rightarrow \rho =1.5625\times { 10 }^{ -5 }\ \Omega m\approx 1.6\times { 10 }^{ -5 }\ \Omega m\)

Concept:

Electron Mobility:

• Electron mobility (μ) is given by the formula: μ = v_d / E, where v_d is the drift speed and E is the electric field.
• The electric field E is related to the potential difference (V) and the length (L) of the conductor: E = V / L.

Calculation:

Given,

Length of the conductor, L = 20 cm = 0.2 m

Potential difference, V = 16 V

Drift speed, v_d = 2.4 × 10^-3 m/s

Electric field, E = V / L = 16 / 0.2 = 80 V/m

Electron mobility, μ = v_d / E = (2.4 × 10^-3) / 80 = 3 × 10^-6 m²/V·s

∴ The electron mobility is 3 × 10^-6 m²/V·s. Option 4) is correct.

The correct answer is Nichrome.

Important Points

• A heating element converts electrical energy into heat by means of the Joule heating mechanism. 
• Nichrome, which consists of 80 per cent nickel and 20 per cent chromium, is typically what most metal heating elements are made of.
• Nichrome 80/20 produces excellent heating elements because of a reasonably high resistance provided by the material. 
• When the heating element is heated for the first time, the chromium alloy reacts with atmospheric oxygen and forms a coating of chromium oxide on the heating element's outer surface.
• This coating of chromium oxide serves as a protective layer for the element and prevents degradation of the material under these layers, preventing the wire of the element from cracking and burning out.
• Heating elements made from Nichrome are suitable for continuous operation at a temperature of up to 1200^oC.

Coil made from Nichrome used in Electric heaters:

The correct answer is option 2) i.e. A/m2

CONCEPT:

• Current density is defined as the amount of electric charge flowing across a unit cross-sectional area in unit time.
  • It is a vector quantity.

The electric current density (J) for a given conducting material is given as:

\(J = \frac{q/t}{A} = \frac{I}{A}\)

Where q is the electric charge flowing, t is the time, I is the current flowing and A is the cross-sectional area.

EXPLANATION:

• The SI units for current and area are A and m² respectively.

Current density, \(J = \frac{I}{A}\) 

Thus, the SI unit of current density is A/m².

CONCEPT:

• Ohms’ law: At constant temperature, the current through a resistance is directly proportional to the potential difference across the resistance.

V = R I

Where V is the potential difference, R is resistance and I is the current flowing

• Current density (J): The electric current per unit area is called current density.

\(Current~density~\left( J \right)=\frac{Current~\left( I \right)}{Area~\left( A \right)}\)

• Conductivity (σ): The property of a conductor due to which the current flows through it is called conductivity of that conductor.

\(Resistance~\left( R \right)=\frac{\rho ~l}{A}\)

Where ρ is resistivity = 1/σ, l is the length and A is the area of the conductor

Electric field (E) = potential difference (V)/length (l)

EXPLANATION:

As V = R I 

\(Resistance~\left( R \right)=\frac{\rho ~l}{A}\)

\(V=\frac{\rho ~l}{A}\times I=\left( \frac{1}{\sigma } \right)\times l\times \frac{I}{A}=\left( \frac{1}{\sigma } \right)\times l\times J\)

\(J=\sigma \times \frac{V}{l}=\sigma ~E\)

Hence option 3 is correct.

The correct answer is Fuse.

• Fuse is a piece of wire of a material with a very low melting point.
• When a high current flows through the circuit due to overloading or a short circuit, the wires get heated and melts.
• As a result, the circuit is broken and current stops flowing.
• The fuse is always connected in the live path because if the live wire gets short-circuited somewhere within the appliance and results in the high current then the fuse can blow and protect the appliance.

Additional Information

• A capacitor is a device that stores electrical energy in an electric field.
  • It is a passive electronic component with two terminals.
  • The effect of a capacitor is known as capacitance. 
• A surge protector is an electrical device that is used to protect equipment against power surges and voltage spikes while blocking voltage over a safe threshold (approximately 120 V). 
• A Protective Relay is a device that detects the fault and initiates the operation of the circuit breaker to isolate the defective element from the rest of the system.

CONCEPT:

• Relaxation time: In a conductor, the time gap between two successive collisions of electrons, when current flows is called the relaxation time.

Relaxation time is calculated by:

\(τ = \frac{m}{ne^2ρ}\)

where m is the mass of the electron, n is the number of free electrons per unit volume, e is the charge on one electron, and ρ is the resistivity.

• Current density: The amount of charge per unit time that flows through a unit area of a chosen cross-section.

J = i/A

where J is current density, i is current and A is cross-section area.

• Drift velocity: In a material, The average velocity attained by charged particles due to an electric field is called drift velocity.

The drift velocity of the electrons is calculated by:

\(v=\frac{I}{neA}\)

where v is the drift velocity, I is the current in the wire, n is the number density of free electrons in the wire, A is the cross-sectional area of the wire, and e is the charge on one electron.

EXPLANATION:

The relation between current density and electric field E:

1st equation of motion

v = u + at

v_d = at

v_d = (eE/m) τ

J = i/A

J = neAv_d/A

J = nevd

J = ne (eE/m) τ

J = ne²Eτ/m

Current density = ne2Eτ/m

So the correct answer is option 3.

The correct answer is 405C.

Key Points

• A filament of an electric bulb carries a current of 0.75 A in 9 minutes. Find the amount of electric charge flowing through the current;
  • ​Current (I): 0.75 A.
  • Time (t): 9 minutes. 
• Time in minutes can be converted into seconds as follows.
  • Time (t) = 9 × 60
  • t = 540 sec
  • Q = I X t
  • Q = 0.75 X 540
  • Q = 405C

Direct Current:

• Direct current (DC) is an electric current that is unidirectional, so the flow of charge is always in the same direction.

• In the direct current (DC), the direction and magnitude of the current do not change. It remains constant with time.
• The positive and negative terminals of a battery are always, respectively, positive and negative.
• Current always flows in the same direction between the terminals. 
• It is used in many household electronics and in all devices that use batteries.
• Examples: Batteries, Fuel cells, solar cells

Direct Current is shown in the figure:

Alternating Current:

• The current that changes its magnitude and polarity at regular intervals of time is called an alternating current.
• The major advantage of using the alternating current instead of direct current is that the alternating current is easily transformed from a higher voltage level to a lower voltage level.
• AC is the most commonly used and most preferred electric power for household equipment, office, and buildings, etc.
• For Alternating current both magnitude and direction change. The frequency of alternating current in Indian power supply is 50 Hz. The time period is 1/50 = 20 m sec.

Alternating Current is shown in the figure

Sinusoidal Current:

• An alternating current that constitutes a sinusoidal function of time in the form I = Im sin (ωt + ϕ)
• The use of sinusoidal currents in technology simplifies electrical devices and circuits, as well as design calculation.

The correct answer is option 2, i.e. Statements 1 and 2 are not correct.

• Lightning is produced by discharges of electricity from cloud to cloud or from cloud to ground. Hence statement 1 is not correct.
• Thunder is the loud noise which follows a flash of lightning.
  • Lightning can be seen before thunder can be heard as light travels faster than sound.
  • Since you see lightning immediately and it takes the sound of thunder about 5 seconds to travel a mile, you can calculate the distance between you and the lightning.
  • If you count the number of seconds between the flash of lightning and the sound of thunder, and then divide by 5, you'll get the distance in miles to the lightning: 5 seconds = 1 mile, 15 seconds = 3 miles, 0 seconds = very close.
  • Hence statement 3 is correct.
• In a thunderstorm, you should not stand under a tree because the light tends to strike the highest point of everything. Hence statement 2 is not correct.

Concept:

The current density for any material is given by:

J = \( {I \over A}\)

where J = current density

I = current 

A = area

Calculation:

Given, J = 9A/cm²

A = 5 cm²

9 = \( {I \over 5}\)

I = 45 A

CONCEPT: 

• ​For the proper functioning of an electrical appliance uninterrupted, the power supply is required, while operating an electrical appliance heat loss occurs as per Joule's law of heating.
• In India, the power supply is achieved through overhead power poles 
• There are three wires in household wiring
  1. The red coloured wire which is known as a live wire, which supplies current to electrical appliances.
  2. Black coloured wire which is known as Neutral wire, which takes back the circuit to the original power supply or else it works as another earthing wire.
  3. Green coloured wire which is known as Earth wire, the function of an earth wire is to bypass the excess current in household wiring to earth in order to protect the electrical appliances. ​

​

EXPLANATION :

• The major aim of household wiring is to supply the proper amount of current to operate the electrical appliances
• From the above, it is clear that Green coloured wire which is known as Earth wire, the function of an earth wire is to bypass the excess current in household wiring to earth in order to protect the electrical appliances. ​
• ​Hence, option 4 is the answer