Electric Current

Which one of the following laws of electromagnetism does not give the direction of magnetic field ?

[amp_mcq option1=”Right-hand thumb rule” option2=”Fleming’s left-hand rule” option3=”Fleming’s right-hand rule” option4=”Faraday’s law of electromagnetic induction” correct=”option4″]

The Right-hand thumb rule gives the direction of the magnetic field produced by a current-carrying wire. Fleming’s Left-hand rule gives the direction of the force acting on a current-carrying conductor placed in a magnetic field (given the directions of current and magnetic field). Fleming’s Right-hand rule gives the direction of the induced current when a conductor moves in a magnetic field (given the directions of motion and magnetic field). Faraday’s law of electromagnetic induction states that a changing magnetic flux through a circuit induces an electromotive force (EMF). While the change in magnetic flux involves the magnetic field, Faraday’s law primarily quantifies the magnitude of the induced EMF and does not directly provide a rule for determining the direction of the magnetic field itself. (Lenz’s law, often used alongside Faraday’s law, provides the direction of the induced current/EMF, opposing the change in flux).

Faraday’s Law quantifies the magnitude of induced EMF due to changing magnetic flux; the other rules/laws listed are primarily used to determine directions related to magnetic fields, current, force, or motion.

Right-hand rules and Fleming’s rules are mnemonic tools for remembering the vector relationships in electromagnetism. Faraday’s law is a fundamental principle linking changing magnetic fields to induced electrical effects.

An electron and a proton starting from rest get accelerated through potential difference of 100 kV. The final speeds of the electron and the proton are V$_e$ and V$_p$ respectively. Which one of the following relations is correct?

[amp_mcq option1=”V$_e$ > V$_p$” option2=”V$_e$ < V$_p$” option3=”V$_e$ = V$_p$” option4=”Cannot be determined” correct=”option1″]

The correct option is A.

When a charged particle is accelerated from rest through a potential difference V, its kinetic energy is equal to the work done on it by the electric field, which is |qV|. So, (1/2)mv^2 = |qV|. The speed is given by v = sqrt(2|q|V/m). The electron and the proton have the same magnitude of charge (|q| = e) and are accelerated through the same potential difference V. Thus, their kinetic energies gained are equal: KE_e = KE_p = eV.
(1/2)m_e * V_e^2 = (1/2)m_p * V_p^2
m_e * V_e^2 = m_p * V_p^2
V_e / V_p = sqrt(m_p / m_e)
Since the mass of the electron (m_e) is significantly less than the mass of the proton (m_p), the ratio m_p / m_e is greater than 1. Therefore, V_e / V_p > 1, which means V_e > V_p. The lighter particle (electron) achieves a much higher speed than the heavier particle (proton) for the same kinetic energy.

The mass of an electron is approximately 9.11 x 10^-31 kg, and the mass of a proton is approximately 1.67 x 10^-27 kg. The proton mass is about 1836 times the electron mass. Thus, the electron’s speed will be about sqrt(1836) ≈ 43 times the proton’s speed.

A wire of copper having length l and area of cross-section A is taken and a current I is flown through it. The power dissipated in the wire is P. If we take an aluminum wire having same dimensions and pass the same current through it, the power dissipated will be

[amp_mcq option1=”P” option2=”< P” option3=”> P” option4=”2P” correct=”option3″]

The power dissipated in a wire is given by P = I²R, where I is the current flowing through the wire and R is its resistance. The resistance of a wire is given by R = ρ * (l/A), where ρ is the resistivity of the material, l is the length, and A is the area of cross-section.
For the copper wire, P = I² * R_copper = I² * (ρ_copper * l / A). We are given this power is P.
For the aluminum wire, it has the same dimensions (l and A) and the same current (I). The power dissipated is P_aluminum = I² * R_aluminum = I² * (ρ_aluminum * l / A).
To compare P_aluminum and P, we need to compare the resistivities of aluminum (ρ_aluminum) and copper (ρ_copper). Copper is a better conductor than aluminum, which means its resistivity is lower. Therefore, ρ_aluminum > ρ_copper.
Since P_aluminum is proportional to ρ_aluminum (for fixed I, l, A) and P is proportional to ρ_copper, and ρ_aluminum > ρ_copper, it follows that P_aluminum > P. The power dissipated in the aluminum wire will be greater than in the copper wire.

– Power dissipated in a resistor is proportional to its resistance (P = I²R, or P = V²/R, or P = VI).
– Resistance depends on the material’s resistivity and the wire’s dimensions (R = ρl/A).
– Aluminum has higher resistivity than copper.

Copper is widely used in electrical wiring due to its low resistivity and ductility. Aluminum is also used, particularly in high-voltage transmission lines, as it is lighter and cheaper than copper, despite having a higher resistivity. The difference in resistivity directly impacts the power loss due to heating (Joule heating) for the same current and dimensions.

Working of safety fuses depends upon

• magnetic effect of the current
• chemical effect of the current
• magnitude of the current
• heating effect of the current

Select the correct answer using the code given below.

[amp_mcq option1=”1, 2, 3 and 4″ option2=”1, 2 and 3 only” option3=”3 and 4 only” option4=”4 only” correct=”option3″]

A safety fuse is designed to protect electrical circuits from overcurrents. It consists of a wire made of a material with a low melting point (like tin-lead alloy) and a specific resistance. When the current flowing through the wire exceeds a safe limit (the fuse rating), the wire heats up significantly due to the **heating effect of the current** (Joule heating, H = I²Rt). If the **magnitude of the current** is sufficiently high, the heat generated melts the fuse wire, breaking the circuit and stopping the flow of current. Therefore, the working of a safety fuse depends on the magnitude of the current and its heating effect.

– The heating effect of electric current states that heat is produced when current flows through a resistance.
– The amount of heat produced is proportional to the square of the current (I²), the resistance (R), and the time (t).
– Fuse wire melts when the heat generated by excessive current raises its temperature to its melting point.
– The specific melting current is determined by the material, length, and thickness of the fuse wire.

Magnetic effect of current is used in devices like circuit breakers which use an electromagnet to trip a switch when current exceeds a limit. Chemical effect of current is associated with electrolysis. Neither magnetic nor chemical effects are the primary working principles of a simple fuse wire melting due to overcurrent. Only the magnitude of current (as it determines the heat) and the heating effect are relevant.

Which one of the following statements is not correct?

[amp_mcq option1=”The cathode rays originate from cathode and proceed towards the anode in a cathode-ray discharge tube.” option2=”The television picture tubes are nothing but cathode-ray tubes.” option3=”The cathode rays themselves are not visible.” option4=”The characteristics of cathode rays depend upon the nature of the gas present in the cathode-ray tube.” correct=”option4″]

Statement D is incorrect. Cathode rays are streams of electrons emitted from the cathode in a discharge tube. The properties of electrons, such as charge and mass, are fundamental constants and do not depend on the material of the cathode or the type of residual gas present in the tube. J.J. Thomson’s experiments showed that the charge-to-mass ratio of cathode rays was always the same, regardless of the source material.

Statements A, B, and C are correct. Cathode rays originate from the negatively charged cathode and travel towards the positively charged anode (A). Older television and computer monitors used Cathode Ray Tubes (CRTs) to produce images (B). The beam of electrons itself is not visible to the naked eye; the light is produced when the electrons strike a fluorescent screen (C).

The discovery and study of cathode rays were crucial in the development of the understanding of the electron as a fundamental particle of matter.

A copper wire of radius r and length 1 has a resistance of R. A second copper wire with radius 2r and length 1 is taken and the two wires are joined in a parallel combination. The resultant resistance of the parallel combination of the two wires will be

[amp_mcq option1=”5R” option2=”5/4 R” option3=”4/5 R” option4=”R/5″ correct=”option4″]

The resistance of a wire is given by the formula R = ρ * (l/A), where ρ is the resistivity of the material, l is the length, and A is the cross-sectional area. The area of a circular wire is A = πr², where r is the radius.

– For the first wire: R₁ = R = ρ * (l / πr²).
– For the second wire: The radius is 2r, and the length is l. So, R₂ = ρ * (l / π(2r)²) = ρ * (l / 4πr²) = (1/4) * [ρ * (l / πr²)] = R/4.
– When two resistors R₁ and R₂ are connected in parallel, the resultant resistance R_eq is given by 1/R_eq = 1/R₁ + 1/R₂ or R_eq = (R₁ * R₂) / (R₁ + R₂).
– Plugging in the values, R_eq = (R * (R/4)) / (R + R/4) = (R²/4) / (5R/4).
– R_eq = (R²/4) * (4/5R) = R/5.

This problem illustrates the relationship between resistance, material properties, and geometry, as well as the calculation of equivalent resistance in a parallel circuit. The resistivity (ρ) is the same for both wires as they are both copper.

Suppose voltage V is applied across a resistance R. The power dissipated in the resistance is P. Now the same voltage V is applied across a parallel combination of three equal resistors each of resistance R. Then the power dissipated in the second case will be

[amp_mcq option1=”P” option2=”3P” option3=”P/3″ option4=”2P/3″ correct=”option2″]

The correct answer is B) 3P.

In the first case, power dissipated is P = V²/R.
In the second case, three equal resistors, each of resistance R, are connected in parallel across the same voltage V. The equivalent resistance of three resistors in parallel is R_eq = 1 / (1/R + 1/R + 1/R) = 1 / (3/R) = R/3.
The power dissipated in the second case is P’ = V²/R_eq. Substituting R_eq = R/3, we get P’ = V²/(R/3) = 3 * (V²/R).
Since P = V²/R, the power dissipated in the second case is P’ = 3P.

When resistors are connected in parallel across a constant voltage source, the total power dissipated is the sum of the power dissipated in each resistor. In this case, each of the three resistors R has voltage V across it, so each dissipates power P = V²/R. The total power is P + P + P = 3P. Using the equivalent resistance method confirms this result.