Current Density: Important Questions

Collegedunia Team logo

Collegedunia Team

Content Curator

Current Density Important Questions and Answers are covered in this article along with a detailed explanation. Current density is the total current which is flowing through one unit of cross-sectional area. If the current flow is uniform, the amount of current that flows through a specific conductor is the same at all points of the conductor, even if the area of the conductor.
Current density formula in a specific portion of the conductor, Current Density (J) = I/A

  • Here, ‘I’ is the amount of current in Ampere
  • ‘A’ is the cross-section area in sq. meters.

Very Short Answer Questions (1 Mark)

Ques. What is the symbol for current density?

Ans. Current Density is represented by the symbol ‘J’. 

Ques. Is the current density scalar or a vector quantity?

Ans. Current density is a vector quantity even though it combines two scalar quantities.  

Ques. Define current density.

Ans. Current density is the amount of charge that flows through a specific cross-sectional area of the conductor. In the case of a steady charge flow, the amount of charge remains constant. Nevertheless, the cross-sectional area differs, which, in turn, leads to different values of current density.

Ques. What is the Unit of Current Density?

Ans. Current Density is measured in Ampere/meter2.

Ques. What is the formula for current density?

Ans. The current density formula is given as – J = I/A

Ques. Define current.

Ans. Current is the flow of electrons from an electrically rich source to an electrical deficit one.

Ques. Write the dimensional formula of Current density.

Ans. Current density, J = I/A

⇒ [J] = [A]/[L2] = [L-2A] or [M0L-2T0A]

Read More:


Short Answer Questions (2 Marks)

Ques. A 10mm2 of copper wire conducts a 2 mA current flow. Determine the current density.

Ans. Current (I) = 2 x 10-3 

A = 10 x 10-3

Therefore, current density (J) = 2 x 10-3/10 x 10-3

J = 0.20 A/m2

Ques. What is the relation between current density and electric field?

Ans. With Ohm’s Law, a connection between the electric field and current density can be determined.

I = neAvd

But, vd = eEτ/m

⇒ I = neA(eEτ/m) = ne2AEτ/m

⇒ I/A = ne2Eτ/m

We know that J = I/A

⇒ J = ne2Eτ/m

Also, conductivity, σ = ne2τ/m

⇒ J = σE

Ques. Find out the current density when 137 Ampere of current flows through a conductor of a 1.2m2 cross-section area.

Ans. Given, I = 137A

A = 1.2m2

We know that, J = I/A

J = 137/1.2

J = 114.66 A/m2

Ques. Find the current density in a wire of diameter 0.2 cm in which a current of 10 A flows.

Ans. Given,

  • Diameter, d = 0.2 cm = 2 x 103 m
  • Current , I = 10 A

Using J = I/A = 4I/πD2 = (4 x 10)/(3.14 x 4 x 10-6)

⇒ J = 3.18 x 106 Am-2

Read More:


Long Answer Questions (3 Marks)

Ques. Define and differentiate between AC and DC Current.

Ans. The differences between AC and DC current can be classified as follows – 

AC Current DC Current
AC charge carriers flow in the opposite direction. DC charge carriers flow in the same direction.
The magnitude of the flow differs with time. The magnitude of the flow differs with time.
The frequency of Alternating Current can vary, however, it is always above zero. The frequency of Direct Current is always zero

Ques. What is the relation between the current density and electron mobility?

Ans. We know that I = neAvd

and current density J = I/A

J = (neAvd)/A

J = nevd

J = ne(eEτ/m)                            [v= eEτ/m]

J = neE(eτ/m)

J = neEμ

Ques. Find out the relation between the current density and the relaxation time.

Ans. Given: I = neAvd

and current density J = I/A

J = (neAvd)/A

J = nevd

J = ne(eEτ/m)                            [As v= eEτ/m]

J = ne2Eτ/m

Ques. Calculate the current density in a uniform wire connected to a battery of 3.5 V EMF and negligible internal resistance. The resistance of a wire is 2.0 Ω with a cross-sectional area of 0.70 x 10-6 m2.

Ans.  Given: Electromotive Force – 3.5 V

Resistance – 2.0 Ω

Area of cross-section = 0.70 x 10-6 m2

Current, I = V/R

I = 3.5/2

I = 1.75 A

Current Density, J = I/A

= 1.75/(0.70 x 10-6 m2)

J = 2.5 x 106 m2

Ques. Find out the current density when 100 Amperes current flows through the battery in a 10 m² area.

Ans. Given: Current I = 100 A,

Area, A = 10 m²

Now, through the current density formula, J = IA

J = 100/10

J = 10 A/m²

Therefore, the current density is 10 A/m².

Ques. A 15 mm² copper wire has 5 mA of current flowing through it. Find its current density.

Ans. Given parameters: Total Current, I = 5 mA

Total Area,  A = 15 mm²

Current density formula, J = I/A

J =  5 × 10−3/15×10−3

J = 0.33 A/m²

Therefore, the current density is 0.33 A/m²


Check out:

CBSE CLASS XII Related Questions

  • 1.
    Two copper wires having their radii in the ratio of 3 : 2 are connected in series across a battery. Find the ratio of the drift velocities of the electrons in the wires.


      • 2.
        Two air-filled capacitors of capacitances $C_1$ and $C_2$ are connected in parallel with a dc battery. After the capacitors are fully charged, a slab of dielectric constant K is inserted between the plates of each capacitor. How will the (i) charge on each capacitor and (ii) energy stored in the capacitor affected after the slab is introduced.


          • 3.
            Read the following paragraph and answer the questions that follow.
            A p-type or n-type semiconductor can be converted into a p-n junction by doping it with suitable impurity. The motion of majority charge carriers causes diffusion current across the junction while the barrier electric field causes motion of minority carriers for drift current. In case of unbiased diode, the diffusion and drift currents are equal. This equilibrium is disturbed by the biasing batteries. Diodes, therefore, allow currents in one direction. This property of diode is used in making rectifiers.


              • 4.
                This ‘average velocity’ is found be few mm/s for currents in range of a few amperes. How then is current established almost the instant a circuit is closed ?


                  • 5.
                    A charged particle $+q$ in an electric field $\vec{E}$ experiences a force in the direction of the electric field. As a result, its kinetic energy changes. Similarly, the charged particle also experiences a force when it moves in a magnetic field $\vec{B}$. But this magnetic force is perpendicular to both velocity $\vec{v}$ of the charged particle and the magnetic field $\vec{B}$, so it cannot change the kinetic energy of the charged particle. Consider two charged particles 1 and 2 of masses $m$ and $\frac{m}{2}$ having charges $-q$ and $+2q$ respectively. They are accelerated from rest through the same potential difference $V$ and acquire kinetic energy $K_1$ and $K_2$. Then they enter in a region of uniform magnetic field $\vec{B}$ perpendicular to their velocities.


                      • 6.
                        Draw a labelled ray diagram of a refracting telescope when it forms image of a distant object at infinity. Derive expression for its magnifying power.

                          CBSE CLASS XII Previous Year Papers

                          Comments


                          No Comments To Show