Electricity and Magnetism Practice Questions — JAMB UTME

Question 1Question →

Match each electromagnetic rule or law on the left with its corresponding physical application or phenomenon on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Fleming's left-hand rule pairs with determining the direction of force on a current-carrying conductor; Right-hand grip rule pairs with determining the direction of magnetic field around a straight wire; Lenz's law pairs with determining the direction of an induced current.

Each rule specifically identifies a vector direction in electromagnetism: Fleming's left-hand rule identifies the force direction on a current-carrying conductor, the Right-hand grip rule gives the magnetic field pattern surrounding a current-carrying wire, and Lenz's law gives the direction of an induced current.

Step-by-Step Solution

1

Identify the primary purpose of Fleming's left-hand rule

It relates magnetic field, current, and motion/force on a current-carrying conductor.

The thumb points to force, index finger to magnetic field, and middle finger to current.

2

Identify the primary purpose of the Right-hand grip rule

It determines the magnetic field direction around a straight wire carrying current.

Grasping the conductor with the thumb pointing in current direction causes the curled fingers to point in the field direction.

3

Identify the primary purpose of Lenz's law

It gives the polarity and direction of induced electromagnetic effects.

Lenz's law ensures conservation of energy by opposing the initial flux change.

Key Concept

Electromagnetic rules and their physical applications

Estimated Time:45s

Question 2Question →

An alternating current (AC) series circuit contains a resistor of resistance $R = 30\ \Omega$ , an inductor of inductive reactance $X_L = 80\ \Omega$ , and a capacitor of capacitive reactance $X_C = 40\ \Omega$ . The circuit is connected across an AC supply with a peak voltage of $V_0 = 100\sqrt{2}\text{ V}$ . What is the average power dissipated in the circuit?

120\text{ W}

240\text{ W}

13.3\text{ W}

400\text{ W}

Show answer & explanation

Answer: $120\text{ W}$

Answer

The average electrical power dissipated in the circuit is

120\text{ W}

.

The average power dissipated in an AC series circuit depends solely on the resistive component. First, converting peak voltage

V_0 = 100\sqrt{2}\text{ V}

yields an RMS voltage of

V_{\text{rms}} = 100\text{ V}

. Computing the circuit impedance yields

Z = \sqrt{30^2 + (80 - 40)^2} = 50\ \Omega

. This produces an RMS current of

I_{\text{rms}} = \frac{100}{50} = 2\text{ A}

. Finally, substituting into

P_{\text{avg}} = I_{\text{rms}}^2 R

gives

(2)^2 \times 30 = 120\text{ W}

.

Step-by-Step Solution

1

Calculate the root-mean-square (RMS) voltage of the AC supply.

V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100\text{ V}

AC power calculations require RMS voltage values rather than peak voltage values.

2

Calculate the net impedance (

Z

) of the series RLC circuit.

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{30^2 + (80 - 40)^2} = \sqrt{30^2 + 40^2} = \sqrt{2500} = 50\ \Omega

Resistance and reactances combine quadratically due to phase differences between voltage across components.

3

Calculate the RMS current (

I_{\text{rms}}

) flowing through the circuit.

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{100\text{ V}}{50\ \Omega} = 2\text{ A}

Ohm's law for AC circuits states

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z}

.

4

Calculate the average power dissipated in the circuit.

P_{\text{avg}} = I_{\text{rms}}^2 R = (2\text{ A})^2 \times 30\ \Omega = 120\text{ W}

Power is dissipated only in resistive elements, as pure inductors and capacitors store and return energy without net loss.

Key Concept

Impedance and Power in Series AC Circuits

Question 3Question →

A cell of electromotive force (e.m.f.) $1.5\text{ V}$ and internal resistance $0.5\text{ }\Omega$ is connected in series with a $2.5\text{ }\Omega$ resistor. What is the terminal potential difference across the cell?

1.25\text{ V}

1.50\text{ V}

0.25\text{ V}

1.75\text{ V}

Show answer & explanation

Answer: $1.25\text{ V}$

Answer

The terminal potential difference across the cell is

1.25\text{ V}

.

When current flows through a closed circuit, voltage drops across both the external load resistor and the cell's internal resistance. Subtracting the lost voltage (

0.25\text{ V}

) from the electromotive force (

1.50\text{ V}

) yields a terminal potential difference of

1.25\text{ V}

.

Step-by-Step Solution

1

Calculate the total resistance of the circuit

R_{\text{total}} = R + r = 2.5\text{ }\Omega + 0.5\text{ }\Omega = 3.0\text{ }\Omega

The external load resistor and the internal resistance of the cell are connected in series.

2

Calculate the total electric current flowing through the circuit

I = \frac{E}{R + r} = \frac{1.5\text{ V}}{3.0\text{ }\Omega} = 0.5\text{ A}

Ohm's law for a complete circuit states that current equals total e.m.f. divided by total circuit resistance.

3

Determine the terminal potential difference across the cell

V = I \times R = 0.5\text{ A} \times 2.5\text{ }\Omega = 1.25\text{ V}

Terminal potential difference is the potential drop across the external resistance.

Key Concept

Terminal Potential Difference and Cell Internal Resistance

Question 4Question →

An electric heater of resistance $20\,\Omega$ carries a current of $5\,\text{A}$ when operated. What is the electrical power dissipated by the heater?

500\,\text{W}

100\,\text{W}

4\,\text{W}

2000\,\text{W}

Show answer & explanation

Answer: $500\,\text{W}$

Answer

500\,\text{W}

According to Joule's law of electric heating, electrical power

P

dissipated in a resistor of resistance

R

carrying current

I

is given by

P = I^2 R

. Substituting the given values

I = 5\,\text{A}

and

R = 20\,\Omega

gives

P = (5\,\text{A})^2 \times 20\,\Omega = 25 \times 20 = 500\,\text{W}

.

Step-by-Step Solution

1

Identify the given quantities from the problem statement

Current

I = 5\,\text{A}

and resistance

R = 20\,\Omega

These parameters are required to calculate the power dissipated.

2

Select the appropriate formula for power in terms of current and resistance

P = I^2 R

Joule's heating law specifies that power dissipation is given by

P = I^2 R

.

3

Substitute the values and evaluate the numerical result

P = (5)^2 \times 20 = 25 \times 20 = 500\,\text{W}

Squaring

5

yields

25

, and multiplying by

20

gives the final power output of

500\,\text{W}

.

Key Concept

Electrical Power Dissipation in a Resistor

Estimated Time:45s

Question 5Question →

An electric current of $2.0\,\text{A}$ flows through a resistor of resistance $5.0\,\Omega$ for a duration of $10.0\,\text{seconds}$ . What is the total electrical energy, in Joules, dissipated by the resistor?

Show answer & explanation

Answer: 200

Answer

The total electrical energy dissipated by the resistor is

200\,\text{J}

.

According to Joule's law of heating, the electrical energy

E

converted into thermal energy when a current

I

flows through a resistor

R

for time

t

is given by

E = I^2 R t

. Substituting

I = 2.0\,\text{A}

,

R = 5.0\,\Omega

, and

t = 10.0\,\text{s}

gives

E = (2.0)^2 \times 5.0 \times 10.0 = 4.0 \times 5.0 \times 10.0 = 200\,\text{J}

.

Step-by-Step Solution

1

Identify the given physical values from the problem statement.

I = 2.0\,\text{A}

,

R = 5.0\,\Omega

,

t = 10.0\,\text{s}

.

These are the necessary parameters to compute electrical energy.

2

State the formula for electrical energy dissipated in a resistor (Joule's law).

E = I^2 R t

Electrical energy is equal to power multiplied by time, where power

P = I^2 R

.

3

Calculate the numerical value of the energy.

E = (2.0)^2 \times 5.0 \times 10.0 = 200\,\text{J}

Squaring the current gives

4.0\,\text{A}^2

, and multiplying by

5.0\,\Omega

and

10.0\,\text{s}

yields

200\,\text{J}

.

Key Concept

Joule's Law of Electrical Heating

Question 6Question →

A $6\text{ }\mu\text{F}$ capacitor is connected in series with a parallel arrangement of a $2\text{ }\mu\text{F}$ capacitor and a $1\text{ }\mu\text{F}$ capacitor. If the entire circuit is connected across a $30\text{ V}$ d.c. power supply, what is the electric charge stored on the $2\text{ }\mu\text{F}$ capacitor?

20\text{ }\mu\text{C}

30\text{ }\mu\text{C}

40\text{ }\mu\text{C}

60\text{ }\mu\text{C}

Show answer & explanation

Answer: $40\text{ }\mu\text{C}$

Answer

The charge stored on the

2\text{ }\mu\text{F}

capacitor is

40\text{ }\mu\text{C}

.

To find the charge on the

2\text{ }\mu\text{F}

capacitor, we first combine the parallel capacitors (

2\text{ }\mu\text{F} + 1\text{ }\mu\text{F} = 3\text{ }\mu\text{F}

). Next, we combine this in series with the

6\text{ }\mu\text{F}

capacitor to get an overall capacitance

C_{eq} = 2\text{ }\mu\text{F}

. The total charge drawn from the

30\text{ V}

supply is

Q = 2\text{ }\mu\text{F} \times 30\text{ V} = 60\text{ }\mu\text{C}

. This total charge enters the parallel combination, giving a potential drop across the parallel branch of

V_p = 60\text{ }\mu\text{C} / 3\text{ }\mu\text{F} = 20\text{ V}

. Therefore, the charge on the

2\text{ }\mu\text{F}

capacitor is

2\text{ }\mu\text{F} \times 20\text{ V} = 40\text{ }\mu\text{C}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the parallel section

C_{p} = 2\text{ }\mu\text{F} + 1\text{ }\mu\text{F} = 3\text{ }\mu\text{F}

Capacitors in parallel add directly.

2

Calculate the total equivalent capacitance of the network

C_{eq} = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2\text{ }\mu\text{F}

The

6\text{ }\mu\text{F}

capacitor and the

3\text{ }\mu\text{F}

parallel equivalent are in series.

3

Find the total charge supplied by the

30\text{ V}

source

Q_{total} = C_{eq} \times V = 2\text{ }\mu\text{F} \times 30\text{ V} = 60\text{ }\mu\text{C}

Total charge is the product of equivalent capacitance and total voltage.

4

Determine the potential difference across the parallel branch

V_{p} = \frac{Q_{total}}{C_{p}} = \frac{60\text{ }\mu\text{C}}{3\text{ }\mu\text{F}} = 20\text{ V}

The total charge flows through the series combination, creating a potential drop across the parallel combination equal to

Q_{total} / C_{p}

.

5

Calculate the charge on the

2\text{ }\mu\text{F}

capacitor

Q_{2\mu\text{F}} = C_{2\mu\text{F}} \times V_{p} = 2\text{ }\mu\text{F} \times 20\text{ V} = 40\text{ }\mu\text{C}

The charge on a specific capacitor in parallel is the product of its capacitance and the voltage across the parallel branch.

Key Concept

Charge and voltage distribution in mixed capacitor networks

Question 7Question →

A uniform metal wire of length $20\,\text{m}$ and total mass $0.034\,\text{kg}$ is manufactured from a material of density $8.5 \times 10^3\,\text{kg/m}^3$ and electrical resistivity $1.7 \times 10^{-8}\,\Omega\cdot\text{m}$ . What is the electrical resistance of the wire in ohms?

Show answer & explanation

Answer: 1.7

Answer

The electrical resistance of the wire is

1.7\,\Omega

.

Combining the density relation

V = \frac{m}{d}

with the geometric expression

V = A \cdot L

gives

A = \frac{m}{d L}

. Substituting this into Pouillet's law

R = \frac{\rho L}{A}

yields

R = \frac{\rho d L^2}{m}

. Evaluating with the given values:

R = \frac{(1.7 \times 10^{-8})(8.5 \times 10^3)(20)^2}{0.034} = 1.7\,\Omega

.

Step-by-Step Solution

1

Calculate the volume of the wire using mass and density

V = 4.0 \times 10^{-6}\,\text{m}^3

Volume is related to mass and density by

V = \frac{m}{d}

.

2

Calculate the cross-sectional area of the wire

A = 2.0 \times 10^{-7}\,\text{m}^2

For a cylindrical wire of uniform cross-section,

V = A \cdot L

, so

A = \frac{V}{L}

.

3

Apply resistivity formula to find electrical resistance

R = 1.7\,\Omega

Resistance is given by

R = \frac{\rho L}{A}

.

Key Concept

Relationship between Resistance, Mass, Density, and Resistivity

Estimated Time:2m 0s

Question 8Question →

An alternating voltage source of $90\text{ V}$ (RMS) is connected in series across a circuit containing a resistor of resistance $12\ \Omega$ , an inductor of reactance $20\ \Omega$ , and a capacitor of reactance $11\ \Omega$ . What is the root-mean-square (RMS) current flowing through the circuit?

2.1\text{ A}

4.3\text{ A}

6.0\text{ A}

7.5\text{ A}

Show answer & explanation

Answer: $6.0\text{ A}$

Answer

The RMS current flowing through the circuit is

6.0\text{ A}

.

The total impedance

Z

of a series RLC circuit is given by

Z = \sqrt{R^2 + (X_L - X_C)^2}

. Substituting the given values gives

Z = \sqrt{12^2 + (20 - 11)^2} = \sqrt{144 + 81} = 15\ \Omega

. Dividing the RMS voltage (

90\text{ V}

) by this impedance yields an RMS current of

6.0\text{ A}

.

Step-by-Step Solution

1

Calculate the net reactance of the series AC circuit

X = X_L - X_C = 20\ \Omega - 11\ \Omega = 9\ \Omega

Inductive and capacitive reactances oppose each other in phase by

180^\circ

.

2

Determine the total impedance (Z) of the series RLC circuit using phasor addition

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\ \Omega

Resistance and net reactance are

90^\circ

out of phase, so impedance is calculated using the Pythagorean theorem.

3

Calculate the RMS current using Ohm's law for AC circuits

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{90\text{ V}}{15\ \Omega} = 6.0\text{ A}

The RMS current is equal to the RMS voltage divided by total circuit impedance.

Key Concept

Impedance and RMS Current in Series RLC Circuits

Question 9Question →

Two cylindrical metallic conductors, $X$ and $Y$ , are connected in series across a direct-current source. Wire $X$ has a diameter of $1.0\,\text{mm}$ and a free-electron density of $6.0 \times 10^{28}\,\text{m}^{-3}$ . Wire $Y$ has a diameter of $3.0\,\text{mm}$ and a free-electron density of $2.0 \times 10^{28}\,\text{m}^{-3}$ . What is the ratio of the drift velocity of free electrons in wire $X$ to that in wire $Y$ ?

3.0

0.33

1.0

9.0

Show answer & explanation

Answer: $3.0$

Answer

The ratio of the drift velocity in wire X to that in wire Y is 3.0.

Because the wires are connected in series, the same current passes through both (

I_X = I_Y

). Using the formula for electric current in terms of drift velocity

I = n A e v_d

, where cross-sectional area

A = \pi d^2 / 4

, we find

v_d \propto 1 / (n d^2)

. Taking the ratio yields

v_X / v_Y = (n_Y d_Y^2) / (n_X d_X^2) = (2.0 \times 10^{28} \times 3.0^2) / (6.0 \times 10^{28} \times 1.0^2) = 18 / 6 = 3.0

.

Step-by-Step Solution

1

Relate electric current to drift velocity and conductor geometry.

I = n A e v_d = n \left( \frac{\pi d^2}{4} \right) e v_d

Electric current

I

depends on free-electron density

n

, cross-sectional area

A

, elementary charge

e

, and electron drift velocity

v_d

.

2

Apply the series connection constraint.

I_X = I_Y \implies n_X d_X^2 v_X = n_Y d_Y^2 v_Y

In a series circuit, the steady current flowing through every conductor is identical.

3

Rearrange to solve for the drift velocity ratio

v_X / v_Y

.

\frac{v_X}{v_Y} = \frac{n_Y d_Y^2}{n_X d_X^2}

Isolating

v_X / v_Y

demonstrates inverse proportionality to electron density and the square of conductor diameter.

4

Substitute the given values into the ratio expression.

\frac{v_X}{v_Y} = \frac{(2.0 \times 10^{28}\,\text{m}^{-3}) \times (3.0\,\text{mm})^2}{(6.0 \times 10^{28}\,\text{m}^{-3}) \times (1.0\,\text{mm})^2} = \frac{2.0 \times 9.0}{6.0 \times 1.0} = \frac{18.0}{6.0} = 3.0

Numerical calculation yields the simplified dimensionless ratio.

Key Concept

Drift Velocity and Current Density in Series Conductors

Question 10Question →

A DC power source with an electromotive force (e.m.f.) of $E = 12.0\text{ V}$ and an internal resistance of $r = 1.0\text{ }\Omega$ is connected to an external load. The load consists of two parallel resistors, $R_1 = 3.0\text{ }\Omega$ and $R_2 = 6.0\text{ }\Omega$ , connected in series with an unknown resistor $R_x$ . A real voltmeter with an internal resistance of $R_v = 90.0\text{ }\Omega$ is placed directly across the terminals of the power source and reads $V = 10.8\text{ V}$ . Calculate the resistance of $R_x$ in ohms.

Show answer & explanation

Answer: 8

Answer

The resistance of

R_x

is

8.0\text{ }\Omega

.

Applying the relationship

V = E - Ir

yields a total current of

1.2\text{ A}

through the cell. The total external resistance connected across the terminals is

9.0\text{ }\Omega

. Subtracting the parallel admittance of the

90.0\text{ }\Omega

voltmeter yields a load resistance of

10.0\text{ }\Omega

. Subtracting the

2.0\text{ }\Omega

equivalent resistance of the parallel pair (

3.0\text{ }\Omega

and

6.0\text{ }\Omega

) gives

8.0\text{ }\Omega

for

R_x

.

Step-by-Step Solution

1

Calculate total current supplied by the cell

I = 1.2\text{ A}

Terminal voltage is related to battery e.m.f. and internal resistance by

V = E - Ir

.

2

Calculate equivalent resistance of the entire external circuit across terminals

R_{\text{ext}} = 9.0\text{ }\Omega

By Ohm's law,

R_{\text{ext}} = \frac{V}{I} = \frac{10.8\text{ V}}{1.2\text{ A}} = 9.0\text{ }\Omega

.

3

Determine the resistance of the main circuit load

R_L

R_L = 10.0\text{ }\Omega

The voltmeter is in parallel with

R_L

, giving

\frac{1}{R_{\text{ext}}} = \frac{1}{R_v} + \frac{1}{R_L}

.

4

Calculate the equivalent resistance

R_p

of the parallel combination of

R_1

and

R_2

R_p = 2.0\text{ }\Omega

R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{3.0 \times 6.0}{3.0 + 6.0} = 2.0\text{ }\Omega

.

5

Determine the value of

R_x

R_x = 8.0\text{ }\Omega

The main circuit load consists of

R_p

in series with

R_x

, so

R_L = R_p + R_x

.

Key Concept

Terminal potential difference, loading effect of measuring instruments, and resistor network analysis

Estimated Time:2m 30s

Question 11Question →

A battery with an electromotive force (e.m.f.) of $12.0\text{ V}$ and an internal resistance of $1.0\text{ }\Omega$ is connected to a circuit containing two resistors, $6.0\text{ }\Omega$ and $3.0\text{ }\Omega$ , connected in parallel with each other. What is the potential difference across the $3.0\text{ }\Omega$ resistor?

8.0\text{ V}

12.0\text{ V}

4.0\text{ V}

3.6\text{ V}

Show answer & explanation

Answer: $8.0\text{ V}$

Answer

The potential difference across the

3.0\text{ }\Omega

resistor is

8.0\text{ V}

.

The parallel external resistors of

6.0\text{ }\Omega

and

3.0\text{ }\Omega

combine to give an equivalent resistance of

2.0\text{ }\Omega

. Including the internal resistance of

1.0\text{ }\Omega

, the total circuit resistance is

3.0\text{ }\Omega

. This leads to a total circuit current of

4.0\text{ A}

. Multiplying this current by the external equivalent resistance of

2.0\text{ }\Omega

gives the terminal voltage of

8.0\text{ V}

, which is identical across both parallel branches.

Step-by-Step Solution

1

Calculate the equivalent resistance (

R_p

) of the parallel combination.

R_p = \frac{6.0 \times 3.0}{6.0 + 3.0} = \frac{18.0}{9.0} = 2.0\text{ }\Omega

Resistors in parallel combine according to reciprocal addition.

2

Find the total circuit resistance (

R_{\text{total}}

) including the internal resistance (

r

).

R_{\text{total}} = R_p + r = 2.0\text{ }\Omega + 1.0\text{ }\Omega = 3.0\text{ }\Omega

The internal resistance of the power supply acts in series with the external equivalent load resistance.

3

Calculate the total current (

I

) drawn from the battery using Ohm's law for a complete circuit.

I = \frac{E}{R_{\text{total}}} = \frac{12.0\text{ V}}{3.0\text{ }\Omega} = 4.0\text{ A}

Current is determined by the total electromotive force divided by the total resistance of the circuit.

4

Determine the terminal potential difference (

V

) across the parallel network.

V = I \times R_p = 4.0\text{ A} \times 2.0\text{ }\Omega = 8.0\text{ V}

Components connected in parallel experience the same potential difference, which equals the terminal voltage of the battery.

Key Concept

Terminal potential difference in circuits with internal resistance

Question 12Question →

A flat circular coil consisting of $100$ turns and having a cross-sectional area of $4.0 \times 10^{-3}\text{ m}^2$ is positioned perpendicular to a uniform magnetic field of $0.50\text{ T}$ . If the direction of the magnetic field is completely reversed in a time interval of $0.02\text{ s}$ , what is the magnitude of the average electromotive force (e.m.f.) induced in the coil?

10\text{ V}

20\text{ V}

0.20\text{ V}

40\text{ V}

Show answer & explanation

Answer: $20\text{ V}$

Answer

20\text{ V}

According to Faraday's law of electromagnetic induction, the magnitude of induced electromotive force equals the rate of change of magnetic flux linkage. Because the magnetic field direction is completely reversed, the change in field magnitude is

\Delta B = 0.50 - (-0.50) = 1.0\text{ T}

. Multiplying by the area (

4.0 \times 10^{-3}\text{ m}^2

) and turn count (

100

), then dividing by the duration (

0.02\text{ s}

) gives

20\text{ V}

.

Step-by-Step Solution

1

Determine the change in magnetic flux density (

\Delta B

)

\Delta B = B_f - B_i = -0.50\text{ T} - 0.50\text{ T} = -1.0\text{ T}

, so

|\Delta B| = 1.0\text{ T}

Reversing the magnetic field direction changes the sign of the flux density relative to the coil area.

2

Calculate the total change in magnetic flux linkage (

\Delta \Phi_{total}

)

\Delta \Phi_{total} = N \cdot A \cdot |\Delta B| = 100 \times (4.0 \times 10^{-3}\text{ m}^2) \times 1.0\text{ T} = 0.40\text{ Wb-turns}

The total magnetic flux linkage is proportional to the number of turns, the surface area, and the net field change.

3

Apply Faraday's law of electromagnetic induction to find the induced e.m.f. (

\mathcal{E}

)

\mathcal{E} = \frac{\Delta \Phi_{total}}{\Delta t} = \frac{0.40\text{ Wb-turns}}{0.02\text{ s}} = 20\text{ V}

The induced e.m.f. magnitude is equal to the rate of change of magnetic flux linkage.

Key Concept

Faraday's Law of Electromagnetic Induction

Question 13Question →

A steady electric current of $2.5\,\text{A}$ flows through a conductor for $4.0\,\text{minutes}$ . What is the total electric charge that passes through any cross-section of the conductor during this period?

10\,\text{C}

96\,\text{C}

600\,\text{C}

150\,\text{C}

Show answer & explanation

Answer: $600\,\text{C}$

Answer

600\,\text{C}

Electric current (

I

) is defined as the rate of charge flow (

Q

) per unit time (

t

), expressed as

Q = I \times t

. Converting

4.0\,\text{minutes}

into seconds gives

4.0 \times 60 = 240\,\text{s}

. Multiplying the current of

2.5\,\text{A}

by

240\,\text{s}

yields

600\,\text{C}

.

Step-by-Step Solution

1

Convert the time duration from minutes to seconds

t = 4.0\,\text{minutes} = 4.0 \times 60\,\text{s} = 240\,\text{seconds}

The standard SI unit for time in electromagnetism formulas is seconds.

2

Apply the formula relating electric charge, current, and time (

Q = I \times t

)

Q = 2.5\,\text{A} \times 240\,\text{s} = 600\,\text{C}

Electric current is defined as the rate of flow of electric charge (

I = \frac{Q}{t}

).

Key Concept

Relationship between Electric Current, Charge, and Time

Question 14Question →

Two $12\text{ }\mu\text{F}$ capacitors are connected in series, and this combination is placed in parallel with a third capacitor of capacitance $6\text{ }\mu\text{F}$ . What is the total equivalent capacitance of the arrangement?

3.0\text{ }\mu\text{F}

4.8\text{ }\mu\text{F}

12.0\text{ }\mu\text{F}

30.0\text{ }\mu\text{F}

Show answer & explanation

Answer: $12.0\text{ }\mu\text{F}$

Answer

The total equivalent capacitance of the network is

12.0\text{ }\mu\text{F}

.

Combining the two identical

12\text{ }\mu\text{F}

capacitors in series gives an equivalent capacitance of

6.0\text{ }\mu\text{F}

. Adding this in parallel with the

6.0\text{ }\mu\text{F}

capacitor yields

6.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 12.0\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance (

C_s

) of the two

12\text{ }\mu\text{F}

capacitors connected in series.

C_s = \frac{12 \times 12}{12 + 12} = 6.0\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}

.

2

Calculate the total equivalent capacitance (

C_{eq}

) by adding

C_s

to the parallel

6.0\text{ }\mu\text{F}

capacitor.

C_{eq} = C_s + C_3 = 6.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 12.0\text{ }\mu\text{F}

Capacitors in parallel add directly:

C_{eq} = C_{series} + C_{parallel}

.

Key Concept

Equivalent capacitance of series-parallel capacitor networks

Question 15Question →

A rectangular coil of $100$ turns with dimensions $0.10\text{ m}$ by $0.20\text{ m}$ is positioned perpendicular to a uniform magnetic field of $0.50\text{ T}$ . The coil is rotated through $90^\circ$ about an axis perpendicular to the field lines in a time interval of $0.040\text{ s}$ , bringing its plane parallel to the magnetic field. What is the magnitude of the average electromotive force (e.m.f.) induced in the coil in volts?

Show answer & explanation

Answer: 25

Answer

The magnitude of the average induced electromotive force in the coil is

25\text{ V}

.

According to Faraday's law, the induced electromotive force magnitude is

E = N \frac{\Delta \Phi}{\Delta t}

. The initial flux through each turn is

\Phi_1 = B A = 0.50\text{ T} \times 0.020\text{ m}^2 = 0.010\text{ Wb}

. When rotated parallel to the field, the final flux

\Phi_2

is

0\text{ Wb}

, so

\Delta \Phi = 0.010\text{ Wb}

. Substituting

N = 100

and

\Delta t = 0.040\text{ s}

yields an induced e.m.f. of

100 \times \frac{0.010}{0.040} = 25\text{ V}

.

Step-by-Step Solution

1

Calculate the cross-sectional area of the rectangular coil.

A = 0.10\text{ m} \times 0.20\text{ m} = 0.020\text{ m}^2

The surface area is required to find the magnetic flux passing through each turn.

2

Determine the change in magnetic flux through one turn of the coil.

Initial flux

\Phi_1 = B A = 0.50 \times 0.020 = 0.010\text{ Wb}

; Final flux

\Phi_2 = 0\text{ Wb}

; Change

\Delta \Phi = 0.010\text{ Wb}

When perpendicular to the magnetic field, maximum flux links the coil. Rotating it parallel reduces the flux linking the coil to zero.

3

Apply Faraday's law of electromagnetic induction to calculate induced e.m.f.

E = N \frac{\Delta \Phi}{\Delta t} = 100 \times \frac{0.010\text{ Wb}}{0.040\text{ s}} = 25\text{ V}

Faraday's law states that the induced e.m.f. magnitude equals the rate of change of total magnetic flux linkage.

Key Concept

Faraday's Law of Electromagnetic Induction

Question 16Question →

An AC series circuit consists of a resistor of resistance $R = 6\ \Omega$ connected in series with an inductor of inductive reactance $X_L = 8\ \Omega$ . What is the total impedance of the circuit?

10\ \Omega

14\ \Omega

2\ \Omega

100\ \Omega

Show answer & explanation

Answer: $10\ \Omega$

Answer

The total impedance of the circuit is

10\ \Omega

.

The net impedance

Z

in an AC series circuit containing a resistor and an inductor is given by

Z = \sqrt{R^2 + X_L^2}

. Substituting

R = 6\ \Omega

and

X_L = 8\ \Omega

yields

Z = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\ \Omega

.

Step-by-Step Solution

1

Identify the formula for impedance in an RL series circuit

Z = \sqrt{R^2 + X_L^2}

In an AC circuit, resistance and inductive reactance add vectorially at a

90^\circ

phase angle.

2

Substitute the given values

R = 6\ \Omega

and

X_L = 8\ \Omega

into the formula

Z = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100}

Squaring each component gives the terms required under the radical.

3

Calculate the square root

Z = 10\ \Omega

The square root of 100 gives the net opposition to current flow (impedance).

Key Concept

Impedance of an RL Series AC Circuit

Estimated Time:45s

Question 17Question →

An electron of mass $9.1 \times 10^{-31}\text{ kg}$ and charge $1.6 \times 10^{-19}\text{ C}$ enters perpendicularly into a uniform magnetic field of flux density $2.0 \times 10^{-3}\text{ T}$ with a speed of $3.2 \times 10^6\text{ m/s}$ . What is the radius of the circular path followed by the electron, expressed in millimeters (mm)?

Show answer & explanation

Answer: 9.1

Answer

The radius of the circular path followed by the electron is 9.1 mm.

When a charge

q

enters a magnetic field

B

perpendicularly at speed

v

, the magnetic force

qvB

supplies the centripetal force

\frac{mv^2}{r}

. Rearranging for radius yields

r = \frac{mv}{qB}

. Substituting

m = 9.1 \times 10^{-31}\text{ kg}

,

v = 3.2 \times 10^6\text{ m/s}

,

q = 1.6 \times 10^{-19}\text{ C}

, and

B = 2.0 \times 10^{-3}\text{ T}

gives

r = 9.1 \times 10^{-3}\text{ m} = 9.1\text{ mm}

.

Step-by-Step Solution

1

Equate the magnetic force to the centripetal force for circular motion

qvB = \frac{mv^2}{r}

A charged particle moving perpendicularly to a magnetic field experiences a magnetic force that acts entirely as a centripetal force.

2

Rearrange the equation to make the orbital radius

r

the subject

r = \frac{mv}{qB}

Cancelling one factor of velocity

v

from both sides allows direct computation of

r

.

3

Substitute the physical values into the formula

r = \frac{(9.1 \times 10^{-31})(3.2 \times 10^6)}{(1.6 \times 10^{-19})(2.0 \times 10^{-3})} = 9.1 \times 10^{-3}\text{ m}

Calculates the radius in standard SI units (meters).

4

Convert the resulting radius from meters to millimeters

r = 9.1 \times 10^{-3}\text{ m} \times 1000\text{ mm/m} = 9.1\text{ mm}

The question explicitly requests the answer in millimeters.

Key Concept

Motion of a charged particle in a uniform magnetic field

Question 18Question →

A straight wire of length $0.50\text{ m}$ carrying a current of $4.0\text{ A}$ is placed perpendicular to a uniform magnetic field of flux density $0.20\text{ T}$ . What is the magnitude of the magnetic force acting on the wire?

0.40\text{ N}

0.80\text{ N}

1.00\text{ N}

0.00\text{ N}

Show answer & explanation

Answer: $0.40\text{ N}$

Answer

The magnetic force acting on the wire is

0.40\text{ N}

.

Using the formula

F = BIL \sin\theta

, when a current-carrying wire is perpendicular to the magnetic field,

\theta = 90^\circ

and

\sin 90^\circ = 1

. Substituting

B = 0.20\text{ T}

,

I = 4.0\text{ A}

, and

L = 0.50\text{ m}

yields

F = 0.20 \times 4.0 \times 0.50 = 0.40\text{ N}

.

Step-by-Step Solution

1

Identify the given values and formula

L = 0.50\text{ m}

,

I = 4.0\text{ A}

,

B = 0.20\text{ T}

,

\theta = 90^\circ

, and

F = B I L \sin\theta

.

The magnetic force on a current-carrying conductor in a uniform magnetic field depends on field strength, current, length, and angle of orientation.

2

Substitute the values into the formula and calculate

F = 0.20 \times 4.0 \times 0.50 \times \sin(90^\circ) = 0.40\text{ N}

.

Since

\sin 90^\circ = 1

, the force is maximum for a given magnetic field and current.

Key Concept

Magnetic force on a current-carrying conductor (

F = BIL \sin\theta

)

Estimated Time:45s

Question 19Question →

A straight conductor of length $0.5\text{ m}$ carries a current of $4.0\text{ A}$ in a uniform magnetic field of flux density $0.20\text{ T}$ . If the conductor is placed at an angle of $30^\circ$ to the magnetic field lines, what is the magnitude of the magnetic force exerted on the conductor?

0.20\text{ N}

0.35\text{ N}

0.40\text{ N}

0.80\text{ N}

Show answer & explanation

Answer: $0.20\text{ N}$

Answer

0.20\text{ N}

The magnetic force on a straight wire carrying current in a uniform magnetic field is given by

F = BIL \sin\theta

. Substituting the given values

B = 0.20\text{ T}

,

I = 4.0\text{ A}

,

L = 0.5\text{ m}

, and

\theta = 30^\circ

yields

F = 0.20 \times 4.0 \times 0.5 \times 0.5 = 0.20\text{ N}

.

Step-by-Step Solution

1

Identify the given physical quantities

L = 0.5\text{ m}

,

I = 4.0\text{ A}

,

B = 0.20\text{ T}

, and

\theta = 30^\circ

.

Extracting values from the problem statement ensures correct substitution.

2

State the magnetic force equation for a straight current-carrying conductor in a magnetic field

F = BIL \sin\theta

Only the component of the magnetic field perpendicular to the current contributes to the magnetic force.

3

Substitute the values and calculate the force magnitude

F = (0.20\text{ T}) \times (4.0\text{ A}) \times (0.5\text{ m}) \times \sin(30^\circ) = 0.20 \times 4.0 \times 0.5 \times 0.5 = 0.20\text{ N}

.

Evaluating

\sin 30^\circ = 0.5

gives the required force value.

Key Concept

Magnetic Force on a Current-Carrying Conductor

Estimated Time:1m 15s

Question 20Question →

Two long parallel straight conductors separated by a distance of $0.20\text{ m}$ in vacuum carry steady currents of $5.0\text{ A}$ and $8.0\text{ A}$ in opposite directions. Given that the permeability of free space $\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$ , what is the magnitude of the magnetic force per unit length acting between the conductors and the nature of this force?

4.0 \times 10^{-5}\text{ N/m}

, repulsive

4.0 \times 10^{-5}\text{ N/m}

, attractive

8.0 \times 10^{-5}\text{ N/m}

, repulsive

2.0 \times 10^{-5}\text{ N/m}

, attractive

Show answer & explanation

Answer: $4.0 \times 10^{-5}\text{ N/m}$ , repulsive

Answer

4.0 \times 10^{-5}\text{ N/m}

, repulsive

Applying the formula for force per unit length gives

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{(4\pi \times 10^{-7})(5.0)(8.0)}{2\pi (0.20)} = 4.0 \times 10^{-5}\text{ N/m}

. Since the currents flow in opposite directions, the magnetic interaction is repulsive.

Step-by-Step Solution

1

Identify given variables and the formula for force per unit length between parallel conductors.

Given:

I_1 = 5.0\text{ A}

,

I_2 = 8.0\text{ A}

,

d = 0.20\text{ m}

,

\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}

. Formula:

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

.

The magnetic force per unit length between two parallel wires depends on the current magnitudes and their distance of separation.

2

Substitute values into the equation and compute the magnitude.

\frac{F}{L} = \frac{(4\pi \times 10^{-7})(5.0)(8.0)}{2\pi (0.20)} = \frac{2 \times 10^{-7} \times 40}{0.20} = 4.0 \times 10^{-5}\text{ N/m}

.

Canceling

4\pi

with

2\pi

yields a multiplier of

2 \times 10^{-7}

, simplifying numerical computation.

3

Determine the direction of force based on current flow direction.

The force is repulsive.

Parallel conductors carrying currents in opposite directions generate magnetic field configurations that force the wires apart.

Key Concept

Magnetic force per unit length between parallel current-carrying conductors

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