Electric Current

111. A metallic wire having length *l* and area of cross-section A has a re

A metallic wire having length *l* and area of cross-section A has a resistance R. It is now connected in series with a wire of same metal but with length 2*l* and area of cross-section 2A. The total resistance of the combination will be:

[amp_mcq option1=”2R” option2=”4R” option3=”R” option4=”R/2″ correct=”option1″]

This question was previously asked in
UPSC Geoscientist – 2020
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The resistance of a metallic 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 area of the cross-section. The first wire has resistance R = ρ * (l / A). The second wire is made of the same metal (same ρ), has length 2l, and area of cross-section 2A. Its resistance R₂ = ρ * (2l / 2A) = ρ * (l / A). Therefore, R₂ = R. When the two wires are connected in series, the total resistance is the sum of individual resistances: R_total = R₁ + R₂ = R + R = 2R.
The resistance of a wire is directly proportional to its length and inversely proportional to its area of cross-section. Resistances in series add up directly.
The resistivity (ρ) is a material property and remains constant for the same metal at the same temperature. In this case, both wires are made of the same metal, so their resistivity is the same.

112. A current carrying circular loop having n number of turns per unit len

A current carrying circular loop having n number of turns per unit length has a current I through it. If the current through it and the number of turns per unit length are doubled, then the magnetic field at the centre of the loop will:

[amp_mcq option1=”remain same.” option2=”increase by four times.” option3=”increase by two times.” option4=”decrease by two times.” correct=”option2″]

This question was previously asked in
UPSC Geoscientist – 2020
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The magnetic field at the center of a current carrying circular loop with N turns and current I is given by the formula B = (μ₀ * N * I) / (2 * R), where R is the radius of the loop. The question states ‘n number of turns per unit length’ which is unusual phrasing for a simple loop, but if interpreted as relating to the total number of turns N, then B is directly proportional to both the number of turns N and the current I (B ∝ N * I). If the number of turns (assuming N = n) is doubled and the current I is also doubled, the new magnetic field B’ will be proportional to (2N) * (2I) = 4 * (N * I). Thus, the magnetic field will increase by four times.
The magnetic field produced by a current-carrying coil (like a loop or solenoid) is directly proportional to the number of turns in the coil and the current flowing through it.
While the phrase “turns per unit length” is typically associated with solenoids or toroids, where the magnetic field inside is B = μ₀ * n * I (with n being turns per unit length), the proportionality B ∝ n * I still holds. Doubling both n (or N) and I will result in the magnetic field increasing by a factor of 2 * 2 = 4, regardless of the specific geometry, as long as the formula is proportional to n*I.

113. Which one of the following statements is correct regarding the magneti

Which one of the following statements is correct regarding the magnetic field inside a current carrying solenoid?

[amp_mcq option1=”The magnetic field inside the solenoid is zero” option2=”The magnetic field inside the solenoid is uniform at all the points” option3=”The magnetic field inside the solenoid increases as we move towards the ends” option4=”The magnetic field inside the solenoid decreases as we move towards the ends” correct=”option2″]

This question was previously asked in
UPSC Geoscientist – 2020
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The correct statement is that the magnetic field inside a current carrying solenoid is uniform at all points (for a sufficiently long solenoid). Option B is correct because, in the ideal case of an infinitely long solenoid, the magnetic field inside is perfectly uniform and parallel to the axis. For a real, long solenoid, this uniformity is a very good approximation over most of the interior volume, except near the ends.
For an ideal, infinitely long solenoid, the magnetic field inside is uniform and given by the formula B = μ₀ * n * I, where μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current. This formula shows the field is constant everywhere inside, independent of the position.
Outside a long solenoid, the magnetic field is nearly zero. Near the ends of a real solenoid, the magnetic field lines spread out, causing the field to become non-uniform and weaker than in the central region. However, in typical questions comparing the field at different points inside, the ideal case (uniform field) or the relative behavior (field strongest in the middle, weaker near ends) is considered. Option B represents the key characteristic of the internal field of a long solenoid.

114. A straight conducting wire carries a DC current through it. Which one

A straight conducting wire carries a DC current through it. Which one of the following statements is true regarding the magnetic field around the wire?

[amp_mcq option1=”The magnetic field does not depend upon the amount of current flowing through the wire” option2=”The magnetic field does not depend upon the distance from the wire” option3=”The magnetic field depends upon the material of the wire” option4=”The magnetic field decreases with an increase from the wire” correct=”option4″]

This question was previously asked in
UPSC Geoscientist – 2020
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The magnetic field around a straight conducting wire carrying a DC current decreases with an increase in distance from the wire.
The magnitude of the magnetic field (B) at a distance ‘r’ from a long, straight wire carrying a current ‘I’ is given by the formula $B = \frac{\mu_0 I}{2\pi r}$ (in vacuum or air). This formula shows that B is directly proportional to the current I and inversely proportional to the distance r.
According to the formula, the magnetic field strength decreases as the distance from the wire increases (option D). The magnetic field strength depends on the current flowing through the wire (option A is false). The magnetic field strength depends on the distance from the wire (option B is false). The formula contains $\mu_0$, the permeability of free space, which is a property of the medium, not the wire material itself (assuming the wire is non-magnetic and the medium is uniform) (option C is false). The direction of the magnetic field lines around the wire is circular, concentric with the wire, and can be determined by the right-hand rule.

115. A charged particle moves through a magnetic field B with a velocity v.

A charged particle moves through a magnetic field B with a velocity v. Which one of the following statements is true for the force (F) experienced by the particle?

[amp_mcq option1=”F is maximum when v and B are parallel to each other.” option2=”F is maximum when v and B are anti-parallel to each other.” option3=”F is maximum when v and B are perpendicular to each other.” option4=”The force F is independent of the angle between v and B.” correct=”option3″]

This question was previously asked in
UPSC CDS-2 – 2023
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The force experienced by a charged particle moving through a magnetic field is given by the Lorentz force formula, F = q(v × B), where q is the charge of the particle, v is its velocity, and B is the magnetic field vector. The magnitude of this force is |F| = |q| |v| |B| sin(θ), where θ is the angle between the velocity vector (v) and the magnetic field vector (B). The force is maximum when sin(θ) is maximum.
The sine function has a maximum value of 1, which occurs when the angle θ is 90 degrees (or π/2 radians). This means the magnetic force is maximum when the velocity of the charged particle is perpendicular to the magnetic field.
When v and B are parallel (θ = 0) or anti-parallel (θ = 180 degrees), sin(θ) = 0, and the magnetic force F is zero. The magnetic force is always perpendicular to both the velocity vector and the magnetic field vector.

116. An electrical circuit having combinations of resistances and capacitan

An electrical circuit having combinations of resistances and capacitance is given below. The current, flowing through the circuit will be

[amp_mcq option1=”1 A” option2=”2 A” option3=”1·5 A” option4=”0·5 A” correct=”option2″]

This question was previously asked in
UPSC CDS-2 – 2022
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The circuit consists of a 10V DC voltage source, a 10 ohm resistor, a 20 ohm resistor, and a 10 µF capacitor. We are asked for the current flowing through the circuit. In a DC circuit with a capacitor, after the capacitor has been fully charged (steady state), no current flows through the branch containing the capacitor.
– The diagram shows the 10 ohm resistor in series with a parallel combination.
– The parallel combination consists of two branches: one with the 20 ohm resistor and the capacitor in series, and another branch which appears to be a simple wire (zero resistance).
– In steady state DC, the capacitor acts as an open circuit, meaning the branch containing the 20 ohm resistor and capacitor has effectively infinite resistance to DC current flow.
– The other branch in parallel has zero resistance (a wire).
– Current always follows the path of least resistance. In parallel, a zero-resistance path effectively shorts out any parallel path with resistance. Thus, all the current passing through the 10 ohm resistor will flow through the wire branch, and no current will flow through the branch containing the 20 ohm resistor and capacitor in steady state.
– The equivalent resistance of the parallel combination is 0 ohms.
– The total resistance of the circuit is the sum of the series resistor and the equivalent parallel resistance: R_total = 10 ohms + 0 ohms = 10 ohms.
– The total current flowing from the battery is given by Ohm’s Law: I = V / R_total = 10V / 10 ohms = 1 A. This is the current flowing through the circuit from the battery.
The 20 ohm resistor and the capacitor only influence the current flow during the transient phase when the capacitor is charging. Once the capacitor is fully charged in DC steady state, its branch becomes effectively open circuit, and the parallel wire branch carries all the current.

117. An electric bulb is connected to a 110 V generator. The current is 0.2

An electric bulb is connected to a 110 V generator. The current is 0.2 A. What is the power of the bulb ?

[amp_mcq option1=”0.22 W” option2=”2.2 W” option3=”22 W” option4=”220 W” correct=”option3″]

This question was previously asked in
UPSC CDS-2 – 2021
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The power (P) of an electrical device is given by the product of the voltage (V) across it and the current (I) flowing through it. Given V = 110 V and I = 0.2 A, the power is P = V × I = 110 V × 0.2 A = 22 Watts.
Electrical power (P) is calculated using the formula P = VI, where V is the voltage in volts and I is the current in amperes. The unit of power is the Watt (W).
Other useful formulas relating power, voltage, current, and resistance (R) are P = I²R and P = V²/R (derived from Ohm’s Law, V=IR). In this problem, only voltage and current were given, so P=VI is the direct formula to use.

118. An electric circuit is consisting of a cell, an ammeter and a nichrome

An electric circuit is consisting of a cell, an ammeter and a nichrome wire of length l. If the length of the wire is reduced to half (l/2), then the ammeter reading

[amp_mcq option1=”decreases to one-half.” option2=”gets doubled.” option3=”decreases to one-third.” option4=”remains unchanged.” correct=”option2″]

This question was previously asked in
UPSC CDS-2 – 2021
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According to Ohm’s law, the current (I) through a circuit is inversely proportional to the resistance (R), assuming the voltage (V) is constant (from the cell). The resistance of a wire is directly proportional to its length (R ∝ l). If the length of the nichrome wire is reduced to half (l/2), its resistance becomes half (R/2). Therefore, the current (I = V/R) will double (I’ = V/(R/2) = 2V/R = 2I). The ammeter reading, which measures the current, will get doubled.
Resistance is directly proportional to length. Current is inversely proportional to resistance.
This assumes the temperature of the wire and other factors remain constant. Nichrome is used as a resistive wire material.

119. The magnetic field lines inside a current carrying long solenoid are i

The magnetic field lines inside a current carrying long solenoid are in the form of

[amp_mcq option1=”ellipse.” option2=”parabola.” option3=”hyperbola.” option4=”parallel straight lines.” correct=”option4″]

This question was previously asked in
UPSC CDS-2 – 2021
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Inside a current-carrying long solenoid, the magnetic field is strong and uniform. The magnetic field lines are parallel to the axis of the solenoid and are close to each other, indicating a uniform field. They are essentially parallel straight lines within the solenoid.
The magnetic field inside an ideal long solenoid is uniform and axial.
Outside the solenoid, the magnetic field is weak and spreads out. The magnetic field lines form closed loops, emerging from one end (North pole) and entering the other end (South pole), similar to a bar magnet, but the field is concentrated and parallel inside.

120. If the speed of a moving magnet inside a coil increases, the electric

If the speed of a moving magnet inside a coil increases, the electric current in the coil

[amp_mcq option1=”increases” option2=”decreases” option3=”reverses” option4=”remains the same” correct=”option1″]

This question was previously asked in
UPSC CDS-2 – 2019
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If the speed of a moving magnet inside a coil increases, the electric current in the coil increases.
– This phenomenon is described by Faraday’s Law of Electromagnetic Induction.
– Faraday’s Law states that the magnitude of the induced electromotive force (EMF) in a circuit is proportional to the rate of change of magnetic flux through the circuit.
– When a magnet moves near a coil, it causes a change in the magnetic flux passing through the coil.
– Increasing the speed of the moving magnet increases the rate at which the magnetic flux changes.
– A greater rate of change of magnetic flux induces a larger EMF in the coil.
– According to Ohm’s Law, a larger induced EMF results in a larger induced current, assuming the resistance of the coil remains constant.
The direction of the induced current is given by Lenz’s Law, which states that the induced current flows in a direction that opposes the change in magnetic flux that produced it. While the direction can reverse if the direction of motion or magnetic field changes, the *magnitude* of the current increases with the speed of the magnet’s movement.
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