Electricity and Magnetism Practice Questions — JAMB UTME — Page 8

Question 141Question →

A dip needle placed within the magnetic meridian at a location on Earth's surface records an angle of dip of $60^\circ$ . If the vertical component of Earth's magnetic field at this location is $3.46 \times 10^{-5}\text{ T}$ , what is the horizontal component of Earth's magnetic field? (Take $\tan 60^\circ = 1.73$ , $\sin 60^\circ = 0.87$ , $\cos 60^\circ = 0.50$ )

2.0 \times 10^{-5}\text{ T}

6.0 \times 10^{-5}\text{ T}

1.73 \times 10^{-5}\text{ T}

4.0 \times 10^{-5}\text{ T}

Show answer & explanation

Answer: $2.0 \times 10^{-5}\text{ T}$

Answer

The horizontal component of Earth's magnetic field is

2.0 \times 10^{-5}\text{ T}

.

The horizontal component

B_h

and vertical component

B_v

of Earth's magnetic field are related by

\tan \theta = \frac{B_v}{B_h}

, where

\theta

is the inclination or dip angle. Substituting

B_v = 3.46 \times 10^{-5}\text{ T}

and

\tan 60^\circ = 1.73

yields

B_h = \frac{3.46 \times 10^{-5}}{1.73} = 2.0 \times 10^{-5}\text{ T}

.

Step-by-Step Solution

1

Identify the relationship between the vertical component (

B_v

), horizontal component (

B_h

), and dip angle (

\theta

).

\tan \theta = \frac{B_v}{B_h}

The angle of dip

\theta

is defined by the direction of Earth's total magnetic field relative to the horizontal plane.

2

Rearrange the equation to express

B_h

in terms of

B_v

and

\tan \theta

.

B_h = \frac{B_v}{\tan \theta}

We need to solve for the horizontal component

B_h

.

3

Substitute the given values

B_v = 3.46 \times 10^{-5}\text{ T}

and

\tan 60^\circ = 1.73

into the equation.

B_h = \frac{3.46 \times 10^{-5}}{1.73} = 2.0 \times 10^{-5}\text{ T}

Performing the numerical division yields the correct horizontal field intensity.

Key Concept

Resolution of Earth's Magnetic Field Components

Question 142Question →

An alternating current (AC) circuit consists of a resistor of resistance $R = 30\ \Omega$ connected in series with a pure inductor across an AC supply of root-mean-square (RMS) voltage $100\ \text{V}$ . If the average power dissipated in the circuit is $120\ \text{W}$ , what is the inductive reactance of the inductor?

20\ \Omega

40\ \Omega

50\ \Omega

70\ \Omega

Show answer & explanation

Answer: $40\ \Omega$

Answer

The inductive reactance of the inductor is

40\ \Omega

.

In an AC circuit containing a resistor and a pure inductor, power is dissipated solely by the resistance. Using

P = I_{\text{rms}}^2 R

, the current is

I_{\text{rms}} = \sqrt{120 / 30} = 2\ \text{A}

. The total impedance

Z

is

V_{\text{rms}} / I_{\text{rms}} = 100 / 2 = 50\ \Omega

. Applying the phasor formula for impedance

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

, solving for

X_L

yields

X_L = \sqrt{50^2 - 30^2} = 40\ \Omega

.

Step-by-Step Solution

1

Calculate the RMS current in the circuit using the average power formula

I_{\text{rms}} = 2\ \text{A}

In an AC circuit with a resistor and a pure inductor, average power is dissipated only by the resistor:

P = I_{\text{rms}}^2 R \implies 120 = I_{\text{rms}}^2 \times 30 \implies I_{\text{rms}}^2 = 4 \implies I_{\text{rms}} = 2\ \text{A}

.

2

Determine the total impedance of the circuit

Z = 50\ \Omega

The total impedance is the ratio of RMS voltage to RMS current:

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

.

3

Calculate the inductive reactance using the impedance relationship for a series RL circuit

X_L = 40\ \Omega

Impedance in a series RL circuit is given by

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

. Substituting the known values gives

50 = \sqrt{30^2 + X_L^2} \implies 2500 = 900 + X_L^2 \implies X_L^2 = 1600 \implies X_L = 40\ \Omega

.

Key Concept

Power Dissipation and Impedance in Series RL AC Circuits

Question 143Question →

A bar magnet is pulled away from a stationary circular coil such that its South pole moves directly away from the front face of the coil. Based on Lenz's law, which polarity is induced on the front face of the coil?

North pole, to attract the receding South pole and oppose its motion

South pole, to repel the receding South pole and oppose its motion

South pole, to attract the receding South pole and assist its motion

North pole, to repel the receding South pole and assist its motion

Show answer & explanation

Answer: North pole, to attract the receding South pole and oppose its motion

Answer

A North pole is induced on the front face of the coil to attract the receding South pole and oppose its motion.

According to Lenz's law, the direction of induced current creates a magnetic field that opposes the change causing it. As the South pole moves away, the decreasing magnetic flux is opposed by an attractive force pulling the magnet back. Therefore, an opposite pole (North pole) is induced on the coil's front face.

Step-by-Step Solution

1

Identify the change causing electromagnetic induction

The South pole of the magnet is moving away from the coil, causing a decrease in magnetic flux through the coil.

Induction is driven by a change in magnetic flux according to Faraday's law.

2

Apply Lenz's law to determine the direction of induced magnetic effect

The induced magnetic field must attempt to pull the magnet back to oppose its withdrawal.

Lenz's law dictates that the direction of an induced current always opposes the motion or change causing it.

3

Determine the required magnetic polarity on the front face

To attract the departing South pole, an opposite magnetic pole (North pole) must be induced on the front face.

Unlike magnetic poles attract each other.

Key Concept

Lenz's Law and Direction of Induced Current

Estimated Time:45s

Question 144Question →

Two long, straight parallel wires separated by a distance of $5.0\text{ cm}$ in air carry equal currents in opposite directions. If the repulsive force per unit length between the wires is $1.6 \times 10^{-3}\text{ N/m}$ , determine the magnitude of the current flowing through each wire in amperes. (Take $\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$ )

Show answer & explanation

Answer: 20

Answer

The magnitude of the current flowing through each wire is

20\text{ A}

.

Using the parallel conductor force formula

\frac{F}{L} = \frac{\mu_0 I^2}{2\pi d}

, we substitute

\frac{F}{L} = 1.6 \times 10^{-3}\text{ N/m}

,

d = 0.05\text{ m}

, and

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

. Simplifying yields

1.6 \times 10^{-3} = 4 \times 10^{-6} I^2

, giving

I^2 = 400

and

I = 20\text{ A}

.

Step-by-Step Solution

1

Recall the expression for force per unit length between two current-carrying parallel wires

\frac{F}{L} = \frac{\mu_0 I^2}{2\pi d}

The magnetic field generated by one wire exerts a magnetic force on the current in the adjacent wire.

2

Convert distance to meters and substitute all given values into the formula

d = 0.05\text{ m}

, leading to

1.6 \times 10^{-3} = \frac{(4\pi \times 10^{-7}) I^2}{2\pi (0.05)}

Standard SI unit for distance is meters, necessary for dimensional consistency.

3

Simplify the equation and compute the current magnitude

1.6 \times 10^{-3} = 4 \times 10^{-6} I^2 \implies I^2 = 400 \implies I = 20\text{ A}

Solving the quadratic term gives the scalar current magnitude in amperes.

Key Concept

Force per unit length between parallel current-carrying conductors

Question 145Question →

At a certain location, the horizontal component of the Earth's magnetic field is $3.0 \times 10^{-5}\text{ T}$ . If an additional uniform horizontal magnetic field of $4.0 \times 10^{-5}\text{ T}$ is applied perpendicular to the magnetic meridian, what is the magnitude of the resultant horizontal magnetic flux density experienced by a compass needle at this location?

1.0 \times 10^{-5}\text{ T}

5.0 \times 10^{-5}\text{ T}

7.0 \times 10^{-5}\text{ T}

1.2 \times 10^{-4}\text{ T}

Show answer & explanation

Answer: $5.0 \times 10^{-5}\text{ T}$

Answer

5.0 \times 10^{-5}\text{ T}

Because the Earth's horizontal field component acts along the magnetic meridian and the external field is applied perpendicular to it, the two fields form a right-angled triangle. Applying Pythagoras' theorem gives

\sqrt{(3.0 \times 10^{-5})^2 + (4.0 \times 10^{-5})^2} = 5.0 \times 10^{-5}\text{ T}

, which represents the true resultant field.

Step-by-Step Solution

1

Identify the vector orientation of the two magnetic fields.

The Earth's horizontal component

B_H

acts along the magnetic meridian (North-South), while the applied field

B_{\text{ext}}

acts perpendicular to it (East-West) at an angle of

\theta = 90^\circ

.

Magnetic flux density is a vector quantity, so direction matters when combining fields.

2

Apply the perpendicular vector addition formula.

B_R = \sqrt{B_H^2 + B_{\text{ext}}^2}

For two vectors acting at right angles (

90^\circ

), the resultant magnitude is given by the Pythagorean theorem.

3

Substitute the given numerical values into the equation.

B_R = \sqrt{(3.0 \times 10^{-5})^2 + (4.0 \times 10^{-5})^2} = \sqrt{(9.0 + 16.0) \times 10^{-10}} = \sqrt{25.0 \times 10^{-10}} = 5.0 \times 10^{-5}\text{ T}

Simplifying the square root yields the exact magnitude of the total horizontal field.

Key Concept

Vector Superposition of Magnetic Fields

Estimated Time:2m 0s

Question 146Question →

An electric water heater operating at a voltage of $240\,\text{V}$ has a heating element of resistance $48\,\Omega$ . It is used to heat $1.5\,\text{kg}$ of water from $20^\circ\text{C}$ to $100^\circ\text{C}$ . If the thermal efficiency of the heating process is $80\%$ , calculate the total electrical energy consumed by the heater in kilojoules ( $\text{kJ}$ ). [Take specific heat capacity of water = $4200\,\text{J}\cdot\text{kg}^{-1}\cdot\text{K}^{-1}$ ]

Show answer & explanation

Answer: 630

Answer

The total electrical energy consumed by the heater is

630\,\text{kJ}

.

The thermal energy required to raise the temperature of

1.5\,\text{kg}

of water by

80^\circ\text{C}

is

Q = 1.5 \times 4200 \times 80 = 504,000\,\text{J} = 504\,\text{kJ}

. Taking into account the

80\%

thermal efficiency, the total electrical energy consumed is

E_{\text{elec}} = \frac{504\,\text{kJ}}{0.80} = 630\,\text{kJ}

.

Step-by-Step Solution

1

Calculate the useful heat energy needed to heat the water.

Q = m c (T_2 - T_1) = 1.5 \times 4200 \times (100 - 20) = 504,000\,\text{J} = 504\,\text{kJ}

.

The thermal energy transferred to the water depends on its mass, specific heat capacity, and temperature increase.

2

Account for the efficiency of the heating element to find total electrical energy input.

E_{\text{elec}} = \frac{Q}{\text{Efficiency}} = \frac{504\,\text{kJ}}{0.80} = 630\,\text{kJ}

.

Since only

80\%

of the electrical energy is converted into useful heat energy for the water, the input electrical energy must be greater than the output heat energy.

Key Concept

Conversion of electrical energy to thermal energy and application of thermal efficiency.

Question 147Question →

An alternating current (AC) circuit operating at a frequency of $50\ \text{Hz}$ contains a resistor of resistance $R = 30\ \Omega$ , an inductor of inductance $L = \frac{0.9}{\pi}\ \text{H}$ , and a capacitor of capacitance $C = \frac{200}{\pi}\ \mu\text{F}$ connected in series. What is the total impedance of the circuit?

50\ \Omega

70\ \Omega

170\ \Omega

40\ \Omega

Show answer & explanation

Answer: $50\ \Omega$

Answer

The total impedance of the AC circuit is

50\ \Omega

.

The inductive reactance is

X_L = 2\pi (50)\left(\frac{0.9}{\pi}\right) = 90\ \Omega

and the capacitive reactance is

X_C = \frac{1}{2\pi (50)\left(\frac{200 \times 10^{-6}}{\pi}\right)} = 50\ \Omega

. Since resistance

R

and net reactance

(X_L - X_C = 40\ \Omega)

are perpendicular vectors in a phasor diagram, the total impedance is calculated using the Pythagorean relation:

Z = \sqrt{30^2 + 40^2} = 50\ \Omega

.

Step-by-Step Solution

1

Calculate the inductive reactance (

X_L

)

X_L = 2\pi f L = 2\pi \times 50 \times \frac{0.9}{\pi} = 90\ \Omega

Inductive reactance depends on supply frequency and inductance.

2

Calculate the capacitive reactance (

X_C

)

X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times \frac{200 \times 10^{-6}}{\pi}} = \frac{1}{0.02} = 50\ \Omega

Capacitive reactance is inversely proportional to supply frequency and capacitance.

3

Determine the net reactance (

X

)

X = X_L - X_C = 90\ \Omega - 50\ \Omega = 40\ \Omega

Inductive and capacitive reactances are

180^\circ

out of phase.

4

Calculate total impedance (

Z

) using phasor addition

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\ \Omega

Resistance and net reactance are

90^\circ

out of phase, requiring right-triangle vector summation.

Key Concept

Total Impedance in a Series RLC AC Circuit

Estimated Time:2m 0s

Question 148Question →

Two long, straight, parallel horizontal conductors are separated vertically by a distance of $2.0\text{ cm}$ . The upper conductor has a mass per unit length of $0.04\text{ kg/m}$ and carries a steady current of $50\text{ A}$ . Assuming the currents in the two conductors flow in opposite directions so that the resulting magnetic force is repulsive, what current (in amperes) must flow through the lower conductor to magnetically levitate and balance the weight of the upper conductor? (Take $g = 9.8\text{ m/s}^2$ and $\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$ )

Show answer & explanation

Answer: 784

Answer

The required current in the lower conductor is

784\text{ A}

.

Equating magnetic repulsion per unit length

\frac{\mu_0 I_1 I_2}{2\pi d}

to weight per unit length

\lambda g

gives

\frac{(2 \times 10^{-7}) \times 50 \times I_2}{0.02} = 0.04 \times 9.8

, which simplifies to

5 \times 10^{-4} I_2 = 0.392

, yielding

I_2 = 784\text{ A}

.

Step-by-Step Solution

1

Equate the upward repulsive magnetic force per unit length to the downward gravitational weight per unit length.

\frac{\mu_0 I_1 I_2}{2\pi d} = \lambda g

For the upper conductor to levitate in vertical static equilibrium, the upward magnetic force per meter must exactly balance its weight per meter.

2

Substitute all given physical values in SI units into the force balance equation.

\frac{(4\pi \times 10^{-7}) \times 50 \times I_2}{2\pi \times 0.02} = 0.04 \times 9.8

Converting distance

d = 2.0\text{ cm} = 0.02\text{ m}

and using mass density

\lambda = 0.04\text{ kg/m}

sets up a single equation with unknown

I_2

.

3

Simplify both sides of the equation.

5 \times 10^{-4} I_2 = 0.392

Calculating

\frac{2 \times 10^{-7} \times 50}{0.02} = 5 \times 10^{-4}\text{ N/(A}\cdot\text{m)}

and

0.04 \times 9.8 = 0.392\text{ N/m}

.

4

Solve for the unknown current

I_2

.

I_2 = \frac{0.392}{5 \times 10^{-4}} = 784\text{ A}

Dividing the weight per meter by the magnetic force coefficient yields the exact required current magnitude.

Key Concept

Interaction force between parallel current-carrying conductors and mechanical equilibrium

Question 149Question →

A dip circle is set up in a vertical plane that is inclined at an angle of $60^\circ$ to the magnetic meridian. The needle comes to rest at an apparent angle of dip of $45^\circ$ . If the actual horizontal component of the Earth's magnetic field in the magnetic meridian is $4.0 \times 10^{-5}\text{ T}$ , what is the vertical component of the Earth's magnetic field at that location?

4.0 \times 10^{-5}\text{ T}

2.0 \times 10^{-5}\text{ T}

3.5 \times 10^{-5}\text{ T}

8.0 \times 10^{-5}\text{ T}

Show answer & explanation

Answer: $2.0 \times 10^{-5}\text{ T}$

Answer

The vertical component of the Earth's magnetic field is

2.0 \times 10^{-5}\text{ T}

.

In a vertical plane inclined at an angle

\alpha

to the magnetic meridian, the vertical component

B_V

remains constant while the effective horizontal component becomes

B_H' = B_H \cos \alpha

. Substituting

B_H = 4.0 \times 10^{-5}\text{ T}

and

\alpha = 60^\circ

yields

B_H' = 2.0 \times 10^{-5}\text{ T}

. Using the formula for apparent dip

\tan \theta' = B_V / B_H'

with

\theta' = 45^\circ

gives

\tan 45^\circ = 1

, which confirms

B_V = B_H' = 2.0 \times 10^{-5}\text{ T}

.

Step-by-Step Solution

1

Determine the effective horizontal component of the magnetic field in the plane of inclination

B_H' = B_H \cos \alpha = (4.0 \times 10^{-5}\text{ T}) \times \cos 60^\circ = 2.0 \times 10^{-5}\text{ T}

When a dip circle is rotated by an angle

\alpha

away from the magnetic meridian, the horizontal component acting along the plane of the dip circle is reduced to

B_H \cos \alpha

.

2

Relate the apparent angle of dip to the vertical component and effective horizontal component

\tan \theta' = \frac{B_V}{B_H'}

The vertical component

B_V

remains unchanged regardless of the vertical plane's orientation.

3

Substitute the given values to solve for

B_V

B_V = B_H' \tan 45^\circ = (2.0 \times 10^{-5}\text{ T}) \times 1 = 2.0 \times 10^{-5}\text{ T}

Since

\tan 45^\circ = 1

, the vertical component is equal to the resolved horizontal component.

Key Concept

Apparent Dip Angle and Resolution of Earth's Magnetic Field Components

Estimated Time:2m 0s

Question 150Question →

A straight horizontal wire of length $0.40\text{ m}$ and mass $0.12\text{ kg}$ carries a steady electric current directed towards the East. The wire is situated in a uniform horizontal magnetic field of $0.60\text{ T}$ directed at an angle of $30^\circ$ North of East. Taking the acceleration due to gravity $g = 10\text{ m s}^{-2}$ , what minimum current must flow through the wire for the magnetic force to act vertically upward and balance the weight of the wire?

5.0\text{ A}

10.0\text{ A}

1.0\text{ A}

5.8\text{ A}

Show answer & explanation

Answer: $10.0\text{ A}$

Answer

10.0\text{ A}

The magnetic force on a current-carrying conductor is given by

F = I L B \sin\theta

. With current flowing East and the magnetic field directed

30^\circ

North of East, the angle

\theta = 30^\circ

. The right-hand rule confirms that

\vec{L} \times \vec{B}

points vertically upward. Setting the upward force equal to the weight

mg

, we get

I \times 0.40 \times 0.60 \times \sin 30^\circ = 0.12 \times 10

, which yields

I = 10.0\text{ A}

.

Step-by-Step Solution

1

Calculate the weight of the horizontal wire

W = mg = 0.12\text{ kg} \times 10\text{ m s}^{-2} = 1.2\text{ N}

The magnetic force must balance the downward gravitational force acting on the wire.

2

Determine the angle

\theta

between the current direction and magnetic field

\theta = 30^\circ

The current flows East and the magnetic field points

30^\circ

North of East, so the angle between the vector length element and magnetic field is

30^\circ

.

3

Express the magnetic force using

F = I L B \sin\theta

F = I \times 0.40\text{ m} \times 0.60\text{ T} \times \sin 30^\circ = 0.12 I

The cross product

\vec{I L} \times \vec{B}

gives the magnitude

I L B \sin\theta

and a vertical upward direction according to the right-hand rule.

4

Equate the magnetic force to the weight and solve for

I

0.12 I = 1.2 \implies I = 10.0\text{ A}

For complete vertical equilibrium, the upward magnetic force must equal the downward weight.

Key Concept

Magnetic force on a current-carrying conductor in a uniform magnetic field

Estimated Time:2m 0s

Question 151Question →

An alternating voltage source described by $V(t) = 210\sqrt{2}\sin(100\pi t)\ \text{V}$ is connected in series with a $40\ \Omega$ resistor, an inductor of inductive reactance $100\ \Omega$ , and a capacitor of capacitive reactance $70\ \Omega$ . What is the root-mean-square (RMS) current flowing through the circuit?

4.2\ \text{A}

1.0\ \text{A}

3.0\ \text{A}

7.0\ \text{A}

Show answer & explanation

Answer: $4.2\ \text{A}$

Answer

The root-mean-square (RMS) current flowing through the circuit is

4.2\ \text{A}

.

The peak voltage is

V_0 = 210\sqrt{2}\ \text{V}

, giving an RMS voltage of

V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = 210\ \text{V}

. The impedance of the series RLC circuit is calculated by

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + (100 - 70)^2} = 50\ \Omega

. Therefore, the RMS current is

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{210}{50} = 4.2\ \text{A}

.

Step-by-Step Solution

1

Determine the RMS voltage from the voltage equation

V_{\text{rms}} = 210\ \text{V}

The standard equation for AC voltage is

V(t) = V_0 \sin(\omega t)

, where

V_0 = 210\sqrt{2}\ \text{V}

. The RMS voltage is

V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = 210\ \text{V}

.

2

Calculate the total impedance of the series RLC circuit

Z = 50\ \Omega

Impedance is determined using phasor addition:

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + (100 - 70)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50\ \Omega

.

3

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

I_{\text{rms}} = 4.2\ \text{A}

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{210\ \text{V}}{50\ \Omega} = 4.2\ \text{A}

.

Key Concept

Impedance and RMS current calculation in series RLC alternating current circuits

Estimated Time:2m 0s

Question 152Question →

Match each physical electromagnetic configuration listed under Column I with its corresponding net magnetic force or motion characteristic listed under Column II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

1. Charged particle moving parallel to magnetic field lines

\rightarrow

Experiences zero magnetic force (

F = 0

) and maintains its initial linear trajectory.
2. Parallel conductors carrying opposite currents

\rightarrow

Experience a mutually repulsive force per unit length given by

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

.
3. Current-carrying rectangular coil parallel to magnetic field

\rightarrow

Experiences maximum net magnetic torque (

\tau = N I A B

) while net translational force is zero.
4. Charged particle injected perpendicularly into magnetic field

\rightarrow

Follows a uniform circular path due to a constant magnetic centripetal force (

F = qvB

).

Each electromagnetic system is correctly paired based on the vector cross-product rules governing magnetic force (

F = q(\mathbf{v} \times \mathbf{B})

and

F = I(\mathbf{L} \times \mathbf{B})

). Parallel motion produces zero force due to zero angle; opposite currents repel per unit length according to Ampère's law; a parallel loop experiences maximal couple/torque without linear displacement; and perpendicular particle velocity results in a constant centripetal deflection into circular motion.

Step-by-Step Solution

1

Analyze the angle

\theta

between velocity and magnetic field for a parallel moving charge

Since velocity is parallel to the field,

\theta = 0^\circ

. The magnetic Lorentz force formula

F = qvB\sin\theta

yields

F = 0

, meaning no deflection occurs.

Magnetic forces require a non-zero perpendicular component of motion relative to the magnetic field direction.

2

Determine the interaction force between parallel conductors with opposite currents

By applying the magnetic field rule for long straight conductors and Fleming's left-hand rule for the resulting force, currents flowing in opposite directions repel each other with force per unit length

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

.

Opposite current directions generate magnetic field lines between the wires that reinforce each other, creating a high-density magnetic field region that pushes the conductors apart.

3

Evaluate forces and torque on a rectangular coil aligned parallel to magnetic field lines

The forces on opposite arms are equal in magnitude and opposite in direction, canceling out net linear translation (

F_{\text{net}} = 0

). However, they act along different lines of action, producing a maximum torque

\tau = N I A B \cos 0^\circ = N I A B

.

Maximum torque occurs when the plane of the coil is parallel to the field because the lever arm for the magnetic forces on the side conductors is at its maximum length.

4

Evaluate the trajectory of a charged particle entering perpendicularly into a magnetic field

At

\theta = 90^\circ

,

\sin 90^\circ = 1

, giving a constant magnetic force

F = qvB

. Because vector force is perpendicular to velocity at every instant, it changes only the direction of velocity, driving the particle into a circular orbit of radius

r = \frac{mv}{qB}

.

A constant magnitude force acting perpendicular to the instantaneous velocity vectors fulfills the exact condition for centripetal acceleration.

Key Concept

Magnetic forces on moving charges and current-carrying conductors under specific geometric alignments and current configurations

Estimated Time:3m 0s

Question 153Question →

A rectangular coil consisting of $50$ turns of wire has dimensions of $0.10\text{ m}$ by $0.05\text{ m}$ and carries a steady electric current of $2.0\text{ A}$ . The coil is suspended in a uniform magnetic field of flux density $0.40\text{ T}$ . If the normal to the plane of the coil makes an angle of $30^\circ$ with the direction of the magnetic field, what is the magnitude of the torque exerted on the coil?

0.10\text{ N}\cdot\text{m}

0.17\text{ N}\cdot\text{m}

0.20\text{ N}\cdot\text{m}

3.00\text{ N}\cdot\text{m}

Show answer & explanation

Answer: $0.10\text{ N}\cdot\text{m}$

Answer

The magnitude of the torque exerted on the coil is

0.10\text{ N}\cdot\text{m}

.

The magnitude of torque on a current-carrying coil situated in a uniform magnetic field is given by

\tau = N I A B \sin\theta

, where

\theta

is the angle between the normal vector of the coil and the magnetic field vector. Substituting

N = 50

,

I = 2.0\text{ A}

,

A = 0.005\text{ m}^2

,

B = 0.40\text{ T}

, and

\theta = 30^\circ

yields

\tau = 50 \times 2.0 \times 0.005 \times 0.40 \times 0.5 = 0.10\text{ N}\cdot\text{m}

.

Step-by-Step Solution

1

Calculate the area

A

of the rectangular coil.

A = \text{length} \times \text{width} = 0.10\text{ m} \times 0.05\text{ m} = 0.005\text{ m}^2

The torque on a coil depends on its total surface area.

2

Identify the given values and formula for magnetic torque.

Formula:

\tau = N I A B \sin\theta

, where

N = 50

,

I = 2.0\text{ A}

,

A = 0.005\text{ m}^2

,

B = 0.40\text{ T}

, and

\theta = 30^\circ

.

When

\theta

is measured relative to the normal of the coil's plane, the sine component determines the effective perpendicular force arm.

3

Substitute the values into the formula and solve for torque

\tau

.

\tau = 50 \times 2.0\text{ A} \times 0.005\text{ m}^2 \times 0.40\text{ T} \times \sin 30^\circ = 100 \times 0.002 \times 0.5 = 0.10\text{ N}\cdot\text{m}

Direct evaluation yields the final magnetic torque.

Key Concept

Torque on a Current-Carrying Loop in a Uniform Magnetic Field

Estimated Time:1m 30s

Question 154Question →

A series alternating current (AC) circuit consists of a resistor of resistance $R = 40\ \Omega$ , an inductor with inductive reactance $X_L = 70\ \Omega$ , and a capacitor with capacitive reactance $X_C = 40\ \Omega$ , connected to an AC voltage supply of $100\ \text{V}$ . What is the power factor of the circuit?

Show answer & explanation

Answer: 0.8

Answer

The power factor of the circuit is 0.8.

The total impedance

Z

of the series circuit is found using

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + (70 - 40)^2} = 50\ \Omega

. The power factor is the ratio of resistance to impedance:

\cos\phi = \frac{R}{Z} = \frac{40}{50} = 0.8

.

Step-by-Step Solution

1

Calculate the net reactance of the circuit.

X = X_L - X_C = 70\ \Omega - 40\ \Omega = 30\ \Omega

In a series RLC circuit, the net reactance is the arithmetic difference between the inductive reactance and the capacitive reactance.

2

Calculate the total impedance of the circuit.

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

Impedance represents the combined opposition to current flow from resistance and net reactance in quadrature.

3

Determine the power factor of the circuit.

\cos\phi = \frac{R}{Z} = \frac{40}{50} = 0.8

The power factor is equal to the cosine of the phase angle, which is defined as the ratio of resistance to total impedance.

Key Concept

Power Factor in AC Circuits

Estimated Time:1m 30s

Question 155Question →

An electric generator supplies power to a workshop through transmission lines having a total resistance of $2\,\Omega$ . The workshop operates electrical equipment drawing a power of $4\,\text{kW}$ at a terminal voltage of $200\,\text{V}$ . What is the total power generated by the generator?

4.8\,\text{kW}

4.0\,\text{kW}

4.04\,\text{kW}

24.0\,\text{kW}

Show answer & explanation

Answer: $4.8\,\text{kW}$

Answer

The total power generated by the generator is

4.8\,\text{kW}

.

The electrical current flowing through the system is

20\,\text{A}

based on the load power of

4000\,\text{W}

at

200\,\text{V}

. The power lost in transmission lines is

P = I^2 R = 20^2 \times 2 = 800\,\text{W}

(

0.8\,\text{kW}

). Therefore, the total electrical power generated by the source must be the sum of the load power and line losses, giving

4.0\,\text{kW} + 0.8\,\text{kW} = 4.8\,\text{kW}

.

Step-by-Step Solution

1

Calculate the current flowing through the circuit using the power and voltage at the workshop.

I = \frac{P_{\text{workshop}}}{V} = \frac{4000\,\text{W}}{200\,\text{V}} = 20\,\text{A}

Current is uniform across the series transmission line.

2

Calculate the power loss dissipated as heat in the transmission lines.

P_{\text{loss}} = I^2 R = (20\,\text{A})^2 \times 2\,\Omega = 400 \times 2 = 800\,\text{W} = 0.8\,\text{kW}

By Joule's law of heating, power dissipated in a resistor carrying current

I

is

I^2 R

.

3

Sum the useful power delivered to the workshop and the transmission power loss to obtain total generated power.

P_{\text{total}} = P_{\text{workshop}} + P_{\text{loss}} = 4.0\,\text{kW} + 0.8\,\text{kW} = 4.8\,\text{kW}

Total energy generated per unit time equals energy consumed by load plus energy lost.

Key Concept

Power Loss in Transmission Lines and Total Source Power

Question 156Question →

A proton of mass $1.67 \times 10^{-27}\text{ kg}$ and charge $1.60 \times 10^{-19}\text{ C}$ enters a uniform magnetic field of flux density $0.50\text{ T}$ at a speed of $4.00 \times 10^6\text{ m s}^{-1}$ . If the path of the proton makes an angle of $30^\circ$ with the magnetic field lines, what is the magnitude of the initial acceleration of the proton?

9.58 \times 10^{13}\text{ m s}^{-2}

1.92 \times 10^{14}\text{ m s}^{-2}

1.66 \times 10^{14}\text{ m s}^{-2}

9.58 \times 10^{12}\text{ m s}^{-2}

Show answer & explanation

Answer: $9.58 \times 10^{13}\text{ m s}^{-2}$

Answer

The magnitude of the initial acceleration of the proton is

9.58 \times 10^{13}\text{ m s}^{-2}

.

The magnetic force on a moving charged particle is given by

F = qvB\sin\theta

. Substituting

q = 1.60 \times 10^{-19}\text{ C}

,

v = 4.00 \times 10^6\text{ m s}^{-1}

,

B = 0.50\text{ T}

, and

\theta = 30^\circ

gives

F = 1.60 \times 10^{-13}\text{ N}

. By Newton's second law, acceleration

a = \frac{F}{m} = \frac{1.60 \times 10^{-13}\text{ N}}{1.67 \times 10^{-27}\text{ kg}} = 9.58 \times 10^{13}\text{ m s}^{-2}

.

Step-by-Step Solution

1

Calculate the magnetic force acting on the proton using

F = qvB\sin\theta

.

F = (1.60 \times 10^{-19}\text{ C}) \times (4.00 \times 10^6\text{ m s}^{-1}) \times (0.50\text{ T}) \times \sin 30^\circ = 1.60 \times 10^{-13}\text{ N}

.

The magnetic force on a moving charge in a magnetic field depends on the component of velocity perpendicular to the field.

2

Apply Newton's second law of motion (

F = ma

) to find acceleration.

a = \frac{F}{m} = \frac{1.60 \times 10^{-13}\text{ N}}{1.67 \times 10^{-27}\text{ kg}} \approx 9.58 \times 10^{13}\text{ m s}^{-2}

.

Dividing the magnetic force by the mass yields the magnitude of the resulting acceleration.

Key Concept

Magnetic force on a moving charged particle and Newton's second law

Question 157Question →

A magnetometer stationed at a field site records the horizontal component of the Earth's magnetic field as $36\ \mu\text{T}$ and the total magnetic field intensity as $45\ \mu\text{T}$ . What is the magnitude of the vertical component of the Earth's magnetic field, in $\mu\text{T}$ ?

Show answer & explanation

Answer: 27

Answer

The magnitude of the vertical component of the Earth's magnetic field is

27\ \mu\text{T}

.

The total magnetic field strength

B

is the hypotenuse of a right-angled triangle formed by the horizontal component

B_h

and vertical component

B_v

. By applying the Pythagorean relation

B^2 = B_h^2 + B_v^2

, substituting

B = 45\ \mu\text{T}

and

B_h = 36\ \mu\text{T}

yields

B_v = \sqrt{45^2 - 36^2} = \sqrt{729} = 27\ \mu\text{T}

.

Step-by-Step Solution

1

Relate total magnetic intensity to its orthogonal components

B^2 = B_h^2 + B_v^2

The total magnetic field vector of the Earth is the vector sum of mutually perpendicular horizontal and vertical components.

2

Isolate the vertical component variable

B_v = \sqrt{B^2 - B_h^2}

Applying the Pythagorean theorem allows direct calculation of the missing perpendicular side.

3

Substitute given values and compute numerical result

B_v = \sqrt{45^2 - 36^2} = \sqrt{2025 - 1296} = \sqrt{729} = 27\ \mu\text{T}

Evaluates the exact magnitude of the vertical component.

Key Concept

Orthogonal resolution of Earth's magnetic field components

Estimated Time:1m 15s

Question 158Question →

At a specific location on the Earth's surface, the horizontal component of the Earth's magnetic field is $3.2 \times 10^{-5}\text{ T}$ and the vertical component is also $3.2 \times 10^{-5}\text{ T}$ . What is the angle of dip at this location?

45^\circ

0^\circ

90^\circ

30^\circ

Show answer & explanation

Answer: $45^\circ$

Answer

The angle of dip at this location is

45^\circ

.

The angle of dip

\theta

is defined by

\tan\theta = \frac{B_v}{B_h}

. Since the vertical and horizontal components of Earth's magnetic field are equal, their ratio is equal to 1. The angle whose tangent is 1 is

45^\circ

.

Step-by-Step Solution

1

Identify the relationship between the horizontal component (

B_h

), vertical component (

B_v

), and angle of dip (

\theta

).

\tan\theta = \frac{B_v}{B_h}

The angle of dip

\theta

is the angle made by the Earth's total magnetic field vector with the horizontal.

2

Substitute the given values into the formula.

\tan\theta = \frac{3.2 \times 10^{-5}\text{ T}}{3.2 \times 10^{-5}\text{ T}} = 1

Both

B_v

and

B_h

are given as

3.2 \times 10^{-5}\text{ T}

.

3

Calculate the angle

\theta

.

\theta = \arctan(1) = 45^\circ

The inverse tangent of 1 is

45^\circ

.

Key Concept

Earth's Magnetic Field Components and Angle of Dip

Question 159Question →

A battery of electromotive force (e.m.f.) $24\,\text{V}$ and internal resistance $2\,\Omega$ is connected across a parallel network consisting of two resistors of resistances $6\,\Omega$ and $12\,\Omega$ . What is the total electrical energy, in Joules, dissipated in the external circuit during an operating time of $5\,\text{minutes}$ ?

Show answer & explanation

Answer: 19200

Answer

The total electrical energy dissipated in the external circuit over 5 minutes is

19,200\,\text{J}

.

To compute the external energy dissipated, first combine the parallel resistors to get an equivalent external resistance of

4\,\Omega

. Adding the

2\,\Omega

internal resistance yields a total circuit resistance of

6\,\Omega

, which draws

4\,\text{A}

of current from the

24\,\text{V}

battery. The power delivered to the external load is

P_{ext} = I^2 R_p = 4^2 \times 4 = 64\,\text{W}

. Multiplying this power by the time duration in seconds (

5 \times 60 = 300\,\text{s}

) yields

19,200\,\text{J}

.

Step-by-Step Solution

1

Calculate the equivalent resistance of the external parallel circuit

R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4\,\Omega

The two external resistors are connected in parallel.

2

Calculate the total current supplied by the battery

I = \frac{E}{R_p + r} = \frac{24}{4 + 2} = \frac{24}{6} = 4\,\text{A}

Ohm's law for a complete circuit accounts for both external resistance and internal resistance.

3

Determine the power dissipated exclusively in the external circuit

P_{ext} = I^2 R_p = (4)^2 \times 4 = 16 \times 4 = 64\,\text{W}

Electrical power dissipated across the external load depends on the total current squared times the equivalent external resistance.

4

Convert time from minutes to seconds and calculate total energy dissipated

t = 5 \times 60 = 300\,\text{s}

,

E = P_{ext} \times t = 64 \times 300 = 19,200\,\text{J}

Electrical energy is the product of power in Watts and time in seconds.

Key Concept

Electrical Energy and Power in Circuits with Internal Resistance

Question 160Question →

Match each electromagnetic rule or concept in Column I with its corresponding physical description or application in Column II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Fleming's Left-Hand Rule matches with determining the direction of magnetic force on a current-carrying conductor; Right-Hand Grip Rule matches with determining the direction of the magnetic field around a straight wire;

F = qvB\sin\theta

matches with calculating the magnitude of magnetic force on a moving electric charge.

Fleming's Left-Hand Rule establishes the perpendicular direction of magnetic force relative to current and magnetic field vectors. The Right-Hand Grip Rule relates linear current direction to circular magnetic field lines. The equation

F = qvB\sin\theta

directly calculates the magnitude of force on a free charge moving through a uniform magnetic field.

Step-by-Step Solution

1

Analyze Fleming's Left-Hand Rule.

The thumb, forefinger, and middle finger represent Force, Magnetic Field, and Current respectively.

It predicts the direction of mechanical motion/force produced in motors and conductors.

2

Analyze the Right-Hand Grip Rule.

Pointing the right thumb in the current's direction makes the curled fingers show the magnetic field line direction.

It provides a simple geometric method to determine field orientation around conductors.

3

Analyze the force relation

F = qvB\sin\theta

.

It quantifies the Lorentz force acting on an isolated moving electric charge.

The magnetic force magnitude depends on charge, velocity, magnetic field strength, and the angle between velocity and field.

Key Concept

Fundamental rules and formulas of electromagnetism

Electricity and Magnetism

Answer

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