Electricity and Magnetism Alıştırma Soruları — JAMB UTME — Sayfa 9

Soru 161Soru →

At a geographical research station, a bar magnet placed along the magnetic meridian neutralizes the horizontal component of the Earth's magnetic field at its neutral points. If the measured horizontal component of the Earth's magnetic field at this site is $1.8 \times 10^{-5}\text{ T}$ and the angle of dip (inclination) is $60^\circ$ , what is the magnitude of the Earth's total magnetic field intensity?

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

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

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

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

Cevabı ve açıklamayı göster

Cevap: $3.6 \times 10^{-5}\text{ T}$

Cevap

The magnitude of the Earth's total magnetic field intensity is

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

.

The horizontal component

B_h

of the Earth's magnetic field is related to the total magnetic intensity

B

and the angle of dip

\theta

by

B_h = B \cos\theta

. Dividing

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

by

\cos 60^\circ = 0.5

gives

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

.

Adım Adım Çözüm

1

Identify given parameters and formula relating field components.

Horizontal component

B_h = 1.8 \times 10^{-5}\text{ T}

, Angle of dip

\theta = 60^\circ

. Formula:

B_h = B \cos\theta

.

The horizontal component of the Earth's magnetic field is the projection of the total field onto the horizontal plane.

2

Rearrange formula to solve for the total magnetic field intensity

B

.

B = \frac{B_h}{\cos\theta}

.

To isolate total magnetic field intensity

B

from the given horizontal component.

3

Substitute numerical values and evaluate.

B = \frac{1.8 \times 10^{-5}\text{ T}}{\cos 60^\circ} = \frac{1.8 \times 10^{-5}}{0.5} = 3.6 \times 10^{-5}\text{ T}

.

\cos 60^\circ = 0.5

, yielding a simple exact calculation.

Anahtar Kavram

Resolution of Earth's Magnetic Field Components

Soru 162Soru →

A straight wire of length $0.50\text{ m}$ carrying a current of $4.0\text{ A}$ is placed in a uniform magnetic field of flux density $0.20\text{ T}$ . If the wire experiences a magnetic force of $0.20\text{ N}$ , what is the angle between the wire and the direction of the magnetic field?

30^\circ

45^\circ

60^\circ

90^\circ

Cevabı ve açıklamayı göster

Cevap: $30^\circ$

Cevap

The angle between the wire and the direction of the magnetic field is

30^\circ

.

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

F = BIL\sin\theta

. Substituting

F = 0.20\text{ N}

,

B = 0.20\text{ T}

,

I = 4.0\text{ A}

, and

L = 0.50\text{ m}

gives

0.20 = 0.40 \sin\theta

, which simplifies to

\sin\theta = 0.50

. Therefore,

\theta = 30^\circ

.

Adım Adım Çözüm

1

Identify the formula for the magnetic force on a current-carrying conductor in a magnetic field.

F = B I L \sin\theta

The magnetic force depends on magnetic flux density

B

, current

I

, length

L

, and the angle

\theta

between the conductor and the magnetic field.

2

Substitute the given values into the magnetic force formula.

0.20 = 0.20 \times 4.0 \times 0.50 \times \sin\theta

Given values are

F = 0.20\text{ N}

,

B = 0.20\text{ T}

,

I = 4.0\text{ A}

, and

L = 0.50\text{ m}

.

3

Solve for

\sin\theta

and calculate

\theta

.

\sin\theta = \frac{0.20}{0.40} = 0.50 \implies \theta = \arcsin(0.50) = 30^\circ

Taking the inverse sine of

0.50

yields an angle of

30^\circ

.

Anahtar Kavram

Magnetic force on a current-carrying conductor (

F = BIL \sin\theta

)

Soru 163Soru →

A flat circular coil of $80$ turns, each having an area of $0.02\text{ m}^2$ , is placed perpendicularly in a uniform magnetic field. If the magnetic flux density decreases uniformly from $0.60\text{ T}$ to $0\text{ T}$ in $0.16\text{ s}$ , what is the magnitude of the induced electromotive force (e.m.f.) in the coil?

0.075\text{ V}

6.0\text{ V}

9.6\text{ V}

15.0\text{ V}

Cevabı ve açıklamayı göster

Cevap: $6.0\text{ V}$

Cevap

6.0\text{ V}

According to Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. is equal to the rate of change of magnetic flux linkage:

\mathcal{E} = N \frac{\Delta \Phi}{\Delta t} = N \cdot A \cdot \frac{\Delta B}{\Delta t}

. Substituting

N = 80

,

A = 0.02\text{ m}^2

,

\Delta B = 0.60\text{ T}

, and

\Delta t = 0.16\text{ s}

yields

\mathcal{E} = 80 \times 0.02 \times \frac{0.60}{0.16} = 6.0\text{ V}

.

Adım Adım Çözüm

1

Calculate the change in magnetic flux density (

\Delta B

) and the change in magnetic flux per turn (

\Delta \Phi

).

\Delta B = 0.60\text{ T} - 0\text{ T} = 0.60\text{ T}

, and

\Delta \Phi = A \times \Delta B = 0.02\text{ m}^2 \times 0.60\text{ T} = 0.012\text{ Wb}

.

Magnetic flux is defined as the product of the perpendicular magnetic flux density and the cross-sectional area.

2

Apply Faraday's Law of Electromagnetic Induction for an

N

-turn coil:

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

.

\mathcal{E} = 80 \times \frac{0.012\text{ Wb}}{0.16\text{ s}} = 80 \times 0.075\text{ V} = 6.0\text{ V}.

The induced e.m.f. is directly proportional to the total rate of change of magnetic flux linkage through all

N

turns of the coil.

Anahtar Kavram

Faraday's Law of Electromagnetic Induction

Soru 164Soru →

When a bar magnet is placed in the magnetic meridian with its North pole pointing towards the Earth's magnetic North pole, the neutral points formed where its magnetic field cancels the horizontal component of the Earth's magnetic field lie along its end-on (axial) line.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: False

Cevap

The statement is false. When a bar magnet is placed with its North pole pointing towards magnetic North, neutral points lie along its broadside-on (equatorial) axis, not its end-on (axial) axis.

The statement is incorrect. When a bar magnet's North pole points North, its magnetic field along its axis points in the same direction as Earth's horizontal field, reinforcing it. The fields oppose each other and form neutral points along the broadside-on (equatorial) line.

Adım Adım Çözüm

1

Determine the direction of Earth's horizontal magnetic field

The horizontal component of Earth's magnetic field (

B_H

) points Northward along the magnetic meridian.

By definition, Earth's horizontal magnetic field vectors point from magnetic South to magnetic North.

2

Analyze the magnetic field of the bar magnet along its axial (end-on) line

Outside the magnet along its axis, field lines point away from the North pole (Northward).

Since both the magnet's field and Earth's horizontal field point Northward along the axial line, they reinforce each other (

B_{\text{total}} = B_{\text{magnet}} + B_H > 0

), so no neutral points can form here.

3

Analyze the magnetic field of the bar magnet along its equatorial (broadside-on) line

Along the equatorial line, field lines curve from the North pole to the South pole, pointing Southward.

Because the magnet's field points Southward while Earth's field points Northward, they oppose each other and cancel when equal in magnitude (

B_{\text{magnet}} = B_H

).

4

Evaluate the statement

The neutral points form on the broadside-on (equatorial) line, making the given statement false.

The statement incorrectly asserts that neutral points occur on the end-on (axial) line.

Anahtar Kavram

Neutral points of a bar magnet in Earth's magnetic field

Tahmini Süre:2m 0s

Soru 165Soru →

A magnetic compass needle free to swing in a vertical plane comes to rest at an angle of dip of $30^\circ$ to the horizontal at a given location. If the magnitude of the Earth's total magnetic field at this point is $5.0 \times 10^{-5}\text{ T}$ , what is the vertical component of the Earth's magnetic field?

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

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

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

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

Cevabı ve açıklamayı göster

Cevap: $2.5 \times 10^{-5}\text{ T}$

Cevap

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

The vertical component of the Earth's magnetic field is found by multiplying the total magnetic field by the sine of the angle of dip:

B_v = B \sin\theta

. Substituting

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

and

\theta = 30^\circ

yields

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

.

Adım Adım Çözüm

1

Identify given parameters and formula for vertical magnetic component

Total magnetic field

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

and angle of dip

\theta = 30^\circ

. Formula:

B_v = B \sin\theta

.

The vertical component

B_v

of the Earth's magnetic field is resolved by projecting the total magnetic flux density

B

along the vertical axis using the sine of the inclination angle.

2

Substitute values and solve for

B_v

B_v = 5.0 \times 10^{-5}\text{ T} \times \sin(30^\circ) = 5.0 \times 10^{-5}\text{ T} \times 0.5 = 2.5 \times 10^{-5}\text{ T}

.

Since

\sin(30^\circ) = 0.5

, multiplying gives the vertical component directly.

Anahtar Kavram

Resolution of Earth's Magnetic Field Components

Tahmini Süre:1m 0s

Soru 166Soru →

At a specific location on Earth, the total magnetic field intensity is $40\ \mu\text{T}$ and the angle of dip is $60^\circ$ . What is the magnitude of the horizontal component of Earth's magnetic field at this location, in $\mu\text{T}$ ?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

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

20\ \mu\text{T}

.

The horizontal component

B_h

of Earth's magnetic field is derived using

B_h = B \cos\theta

. Substituting

B = 40\ \mu\text{T}

and

\theta = 60^\circ

yields

B_h = 40 \times 0.5 = 20\ \mu\text{T}

.

Adım Adım Çözüm

1

Recall the resolving formula for the horizontal component of Earth's magnetic field

B_h = B \cos\theta

The horizontal component is the vector projection of total field

B

onto the horizontal plane inclined at angle

\theta

.

2

Evaluate the cosine function for

60^\circ

\cos(60^\circ) = 0.5

Standard trigonometric value for

60^\circ

.

3

Multiply total magnetic field strength by

\cos(60^\circ)

B_h = 40 \times 0.5 = 20\ \mu\text{T}

Calculates the exact horizontal component magnitude.

Anahtar Kavram

Components of Earth's Magnetic Field

Soru 167Soru →

A short bar magnet with magnetic dipole moment $1.6\text{ A}\cdot\text{m}^2$ is placed along the magnetic meridian with its north pole pointing towards the Earth's magnetic south pole. A neutral point is located on the axial line of the magnet at a distance of $0.2\text{ m}$ from its center. What is the magnitude of the horizontal component of the Earth's magnetic field at this location, in microtesla ( $\mu\text{T}$ )? (Take $\frac{\mu_0}{4\pi} = 10^{-7}\text{ T}\cdot\text{m}\cdot\text{A}^{-1}$ )

Cevabı ve açıklamayı göster

Cevap: 40

Cevap

The magnitude of the horizontal component of Earth's magnetic field at this location is 40 μT.

At a neutral point, the horizontal component of Earth's magnetic field is equal in magnitude and opposite in direction to the magnetic field generated by the bar magnet. For a magnet aligned with its north pole pointing south, neutral points lie on its axial line at distance

d

. Using

B_h = \frac{\mu_0}{4\pi} \frac{2M}{d^3}

with

M = 1.6\text{ A}\cdot\text{m}^2

and

d = 0.2\text{ m}

yields

B_h = 4.0 \times 10^{-5}\text{ T} = 40\ \mu\text{T}

.

Adım Adım Çözüm

1

Determine the condition for the neutral point

B_{\text{axial}} = B_h

When a magnet's north pole points south, its axial magnetic field opposes Earth's horizontal field, creating neutral points along the axis where the magnetic fields cancel out completely.

2

Apply the short bar magnet formula for field along the axial line

B_h = \frac{\mu_0}{4\pi} \frac{2M}{d^3}

The magnetic field produced at an axial point at distance

d

from the center of a short bar magnet of magnetic moment

M

is given by this formula.

3

Substitute the given numerical parameters

B_h = 10^{-7} \times \frac{2 \times 1.6}{(0.2)^3}

Substituting

M = 1.6\text{ A}\cdot\text{m}^2

,

d = 0.2\text{ m}

, and

\frac{\mu_0}{4\pi} = 10^{-7}\text{ T}\cdot\text{m}\cdot\text{A}^{-1}

into the field equation.

4

Calculate the magnitude of the horizontal field component in microtesla

B_h = 4.0 \times 10^{-5}\text{ T} = 40\ \mu\text{T}

Dividing

3.2 \times 10^{-7}

by

8 \times 10^{-3}

gives

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

, which converts to

40\ \mu\text{T}

.

Anahtar Kavram

Neutral points created by a bar magnet aligned with Earth's magnetic meridian

Soru 168Soru →

Two long, parallel straight wires separated by a distance of $0.10\text{ m}$ carry currents of $5.0\text{ A}$ and $8.0\text{ A}$ in opposite directions. What is the magnitude and nature of the magnetic force per unit length exerted between the two wires? (Take $\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}$ )

8.0 \times 10^{-5}\text{ N m}^{-1}

, repulsive

8.0 \times 10^{-5}\text{ N m}^{-1}

, attractive

4.0 \times 10^{-5}\text{ N m}^{-1}

, repulsive

4.0 \times 10^{-5}\text{ N m}^{-1}

, attractive

Cevabı ve açıklamayı göster

Cevap: $8.0 \times 10^{-5}\text{ N m}^{-1}$ , repulsive

Cevap

The magnetic force per unit length is

8.0 \times 10^{-5}\text{ N m}^{-1}

and the nature of the force is repulsive.

The magnetic force per unit length between two parallel conductors carrying currents

I_1

and

I_2

separated by distance

d

is given by

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

. Substituting

I_1 = 5.0\text{ A}

,

I_2 = 8.0\text{ A}

,

d = 0.10\text{ m}

, and

\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}

yields

\frac{F}{L} = 8.0 \times 10^{-5}\text{ N m}^{-1}

. Furthermore, conductors carrying currents in opposite directions exert a repulsive force on each other.

Adım Adım Çözüm

1

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

Given

I_1 = 5.0\text{ A}

,

I_2 = 8.0\text{ A}

,

d = 0.10\text{ m}

, and

\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}

. Formula:

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

.

The magnetic field produced by one current-carrying conductor exerts a force on the second parallel conductor.

2

Substitute the values into the formula and evaluate the magnitude of the force per unit length.

$\frac{F}{L} = \frac{(4\pi \times 10^{-7}) \times 5.0 \times 8.0}{2\pi \times 0.10} = \frac{2 \times 10^{-7} \times 40.0}{0.10} = 8.0 \times 10^{-5}\text{ N m}^{-1}$.

Simplifying

\frac{4\pi}{2\pi} = 2

allows direct scalar computation.

3

Determine the direction/nature of the magnetic force based on current direction.

Since the currents flow in opposite directions, the magnetic forces between the conductors are repulsive.

By the right-hand grip rule and magnetic force laws, parallel currents in opposite directions experience mutual repulsion.

Anahtar Kavram

Magnetic force per unit length between parallel current-carrying conductors

Soru 169Soru →

Match each physical electromagnetic scenario in Column I with its corresponding motion or force behavior in Column II.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Stationary charge matches zero magnetic force; Perpendicular moving charge matches circular trajectory; Positively charged particle moving North in a downward magnetic field matches deflection toward the West; Parallel conductors carrying currents in the same direction match mutual attractive force.

Each physical scenario strictly adheres to fundamental electromagnetic principles: static charges experience no magnetic force, perpendicular motion creates centripetal acceleration leading to circular paths, vector cross products govern particle deflection, and like parallel currents attract.

Adım Adım Çözüm

1

Evaluate the magnetic force on a stationary charge.

Since

v = 0

,

F = q(0)B\sin\theta = 0\text{ N}

.

Magnetic fields only exert forces on moving charges.

2

Determine the path of a charge moving perpendicular to a uniform magnetic field.

The force is perpendicular to velocity at all points, changing direction without changing speed.

A constant perpendicular force acts as a centripetal force, resulting in a circular orbit.

3

Determine force direction for a charge moving North in a downward field.

The vector cross product of North (forward) and Downward yields West.

The direction of magnetic force on a positive charge follows

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

.

4

Determine force interaction between parallel conductors with like currents.

The magnetic field from each wire exerts an inward force on the other wire.

Currents flowing in the same direction attract, while opposite currents repel.

Anahtar Kavram

Magnetic force on moving charges and current-carrying conductors

Soru 170Soru →

A proton carrying a charge of $1.6 \times 10^{-19}\text{ C}$ enters a uniform magnetic field of flux density $0.50\text{ T}$ perpendicularly at a speed of $2.0 \times 10^{6}\text{ m/s}$ . What is the magnitude of the magnetic force acting on the proton?

1.6 \times 10^{-13}\text{ N}

3.2 \times 10^{-13}\text{ N}

6.4 \times 10^{-13}\text{ N}

1.67 \times 10^{-21}\text{ N}

Cevabı ve açıklamayı göster

Cevap: $1.6 \times 10^{-13}\text{ N}$

Cevap

The magnetic force acting on the proton is

1.6 \times 10^{-13}\text{ N}

.

The magnitude of the force on a charge moving through a magnetic field is given by

F = qvB\sin\theta

. Substituting

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

,

v = 2.0 \times 10^6\text{ m/s}

,

B = 0.50\text{ T}

, and

\theta = 90^\circ

produces

1.6 \times 10^{-13}\text{ N}

.

Adım Adım Çözüm

1

Identify the given physical quantities and formula

Charge

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

, velocity

v = 2.0 \times 10^{6}\text{ m/s}

, field strength

B = 0.50\text{ T}

, and angle

\theta = 90^\circ

. The formula for magnetic force on a moving charge is

F = qvB\sin\theta

.

The magnetic force on a moving charged particle depends on charge magnitude, velocity, magnetic flux density, and the angle between velocity and field vectors.

2

Substitute the values into the force equation

F = (1.6 \times 10^{-19}) \times (2.0 \times 10^{6}) \times 0.50 \times \sin(90^\circ) = 1.6 \times 10^{-13}\text{ N}

.

Since

\sin(90^\circ) = 1

, evaluating the product yields the magnetic force in newtons.

Anahtar Kavram

Magnetic Force on a Moving Charge

Soru 171Soru →

An alpha particle carrying a positive charge of $+3.2 \times 10^{-19}\text{ C}$ moves horizontally along the $+x$ -axis at a velocity of $5.0 \times 10^6\text{ m/s}$ . It enters a velocity selector region containing a uniform electric field of $4.0 \times 10^4\text{ V/m}$ directed along the $+y$ -axis. What magnitude and direction of uniform magnetic field $\vec{B}$ are required for the particle to traverse the region undeflected?

8.0 \times 10^{-3}\text{ T}

directed along the

+z

-axis (out of the page)

8.0 \times 10^{-3}\text{ T}

directed along the

-z

-axis (into the page)

2.0 \times 10^{11}\text{ T}

directed along the

+z

-axis (out of the page)

1.25 \times 10^{-2}\text{ T}

directed along the

-y

-axis

Cevabı ve açıklamayı göster

Cevap: $8.0 \times 10^{-3}\text{ T}$ directed along the $+z$ -axis (out of the page)

Cevap

The required magnetic field has a magnitude of

8.0 \times 10^{-3}\text{ T}

directed along the

+z

-axis (out of the page).

The condition for undeflected movement through crossed electric and magnetic fields (velocity selector) requires the magnetic force to be equal and opposite to the electric force. The magnitude is

B = E / v = (4.0 \times 10^4) / (5.0 \times 10^6) = 8.0 \times 10^{-3}\text{ T}

. For a positively charged particle moving along the

+x

-axis with an electric force in the

+y

-direction, Fleming's left-hand rule (or right-hand vector cross product) specifies that the magnetic field must point out of the page along the

+z

-axis to direct the magnetic force along the

-y

-direction.

Adım Adım Çözüm

1

Set up the force balance condition for undeflected motion.

Net force

F_{net} = F_E - F_B = 0 \implies qE = qvB \sin\theta

. Since

\vec{v}

and

\vec{B}

are perpendicular,

\sin\theta = 1

, giving

qE = qvB

.

For zero deflection, the electric force and magnetic force must be equal in magnitude and opposite in direction.

2

Calculate the magnitude of the magnetic field

B

.

B = \frac{E}{v} = \frac{4.0 \times 10^4\text{ V/m}}{5.0 \times 10^6\text{ m/s}} = 8.0 \times 10^{-3}\text{ T}

.

Dividing electric field strength by particle speed yields the required magnetic field strength.

3

Determine the direction of the magnetic field using vector cross-product rules.

Electric force

\vec{F}_E = q\vec{E}

points in the

+y

-direction. Thus, magnetic force

\vec{F}_B = q(\vec{v} \times \vec{B})

must point in the

-y

-direction. With

\vec{v}

in the

+x

-direction (

\hat{i}

),

\hat{i} \times \hat{k} = -\hat{j}

, so

\vec{B}

must point along the

+z

-axis (out of the page).

Opposing vector directions are necessary to achieve equilibrium between electric and magnetic forces.

Anahtar Kavram

Velocity Selector and Lorentz Force Equilibrium

Tahmini Süre:2m 0s

Soru 172Soru →

An electron carrying a charge of magnitude $1.6 \times 10^{-19}\text{ C}$ moves with a velocity of $4.0 \times 10^6\text{ m s}^{-1}$ at an angle of $30^\circ$ relative to a uniform magnetic field. If the magnetic force acting on the electron is $3.2 \times 10^{-13}\text{ N}$ , what is the magnetic flux density of the field?

0.5\text{ T}

1.0\text{ T}

2.0\text{ T}

0.25\text{ T}

Cevabı ve açıklamayı göster

Cevap: $1.0\text{ T}$

Cevap

The magnetic flux density of the field is

1.0\text{ T}

.

Applying the Lorentz force equation for a charged particle

F = qvB\sin\theta

, substituting the given values yields

3.2 \times 10^{-13} = (1.6 \times 10^{-19})(4.0 \times 10^6) B \sin(30^\circ)

. Solving for

B

gives

B = 1.0\text{ T}

.

Adım Adım Çözüm

1

Identify the formula for magnetic force on a moving charge

F = qvB\sin\theta

A charged particle moving at an angle through a magnetic field experiences a force proportional to the velocity component perpendicular to the field.

2

Substitute the given physical values into the equation

3.2 \times 10^{-13} = (1.6 \times 10^{-19}) \times (4.0 \times 10^6) \times B \times \sin(30^\circ)

Values provided: charge

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

, velocity

v = 4.0 \times 10^6\text{ m s}^{-1}

, force

F = 3.2 \times 10^{-13}\text{ N}

, and angle

\theta = 30^\circ

.

3

Evaluate

\sin(30^\circ)

and simplify the numerical product

3.2 \times 10^{-13} = (1.6 \times 10^{-19}) \times (4.0 \times 10^6) \times 0.5 \times B = 3.2 \times 10^{-13} \times B

Since

\sin(30^\circ) = 0.5

, multiplying

1.6 \times 10^{-19} \times 4.0 \times 10^6 \times 0.5

yields

3.2 \times 10^{-13}

.

4

Solve for the magnetic flux density

B

B = \frac{3.2 \times 10^{-13}}{3.2 \times 10^{-13}} = 1.0\text{ T}

Dividing both sides of the equation by

3.2 \times 10^{-13}\text{ N C}^{-1}\text{ m}^{-1}\text{ s}

yields the magnetic flux density.

Anahtar Kavram

Magnetic Force on a Moving Charge

Soru 173Soru →

A long straight conductor lying in the plane of the page carries a steady current directed towards the top of the page. A rectangular conducting loop lies in the same plane to the right of the conductor. If the loop is pulled horizontally to the right away from the conductor, what is the direction of the induced current in the loop and the direction of the net magnetic force exerted on the loop?

Clockwise induced current and a net magnetic force directed to the left (towards the conductor)

Counter-clockwise induced current and a net magnetic force directed to the right (away from the conductor)

Clockwise induced current and a net magnetic force directed to the right (away from the conductor)

Counter-clockwise induced current and a net magnetic force directed to the left (towards the conductor)

Cevabı ve açıklamayı göster

Cevap: Clockwise induced current and a net magnetic force directed to the left (towards the conductor)

Cevap

The induced current flows in a clockwise direction, and the net magnetic force acts to the left (towards the straight conductor).

The straight wire creates a magnetic field pointing into the page on its right side. Pulling the loop further away decreases the inward magnetic flux passing through it. By Lenz's law, the induced current must create an inward magnetic field to oppose this decrease, which corresponds to a clockwise current. Furthermore, Lenz's law requires the resulting mechanical magnetic force to oppose the rightward motion, producing a net force directed to the left (towards the conductor).

Adım Adım Çözüm

1

Determine magnetic field direction around the long straight wire

Using the right-hand grip rule, the magnetic field

\vec{B}

produced by the upward current to the right of the wire points perpendicularly into the page.

Current flowing upward creates concentric magnetic field lines that enter the plane of the page on the right side.

2

Analyze magnetic flux change when moving the loop away

As the loop moves rightward away from the wire, it enters a region of weaker magnetic field, so magnetic flux pointing into the page decreases.

Magnetic field strength decreases inversely with distance (

B \propto 1/r

).

3

Apply Lenz's Law to find induced current direction

The induced current must oppose the decrease in magnetic flux into the page by producing its own magnetic field directed into the page. By the right-hand rule for loops, a clockwise current produces an inward field.

Lenz's law states that an induced current always flows in a direction such that its magnetic field opposes the change in magnetic flux causing it.

4

Determine the net magnetic force on the loop

According to Lenz's law, the net force must oppose the motion causing the induction. Since the loop moves right, the net force must point left (towards the conductor).

Alternatively, the left edge of the loop is closer to the wire and carries upward current (clockwise loop), parallel to the main wire's upward current. Parallel currents attract, yielding a net force to the left.

Anahtar Kavram

Lenz's Law and Electromagnetic Induction

Alternatif Yöntem

Consider magnetic forces on parallel current segments: The left side of the rectangular loop has current flowing upward (for clockwise flow), which is parallel to the main wire's upward current and is therefore attracted to the left. The right side has downward current (antiparallel) and is repelled to the right. Because the left side is closer to the wire, the attractive force dominates, yielding a net force to the left.

Tahmini Süre:1m 15s

Soru 174Soru →

When a soft-iron core is inserted into an air-core solenoid connected to a constant-voltage alternating current (AC) source, the root-mean-square (rms) current flowing through the circuit increases.

Doğru

Yanlış

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Cevap: False

Cevap

The statement is False. Inserting a soft-iron core increases the self-inductance and inductive reactance of the coil, which decreases the rms current.

Inserting a soft-iron core raises the magnetic permeability, thereby increasing the coil's self-inductance (

L

) and inductive reactance (

X_L = 2\pi f L

). This higher reactance opposes current flow more strongly, causing the rms current to drop.

Adım Adım Çözüm

1

Determine the effect of a soft-iron core on self-inductance.

Soft iron has high magnetic permeability, which concentrates magnetic flux lines and markedly increases the coil's self-inductance (

L

).

Self-inductance is directly proportional to the magnetic permeability of the core material.

2

Relate self-inductance to inductive reactance in an AC circuit.

Inductive reactance is given by

X_L = 2\pi f L

. An increase in

L

produces a proportional increase in

X_L

.

Inductive reactance represents the opposition offered by an inductor to alternating current.

3

Calculate the impact on root-mean-square (rms) current.

Using Ohm's law for AC reactive circuits,

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

. Since

X_L

increases,

I_{\text{rms}}

decreases.

Current is inversely proportional to reactance for a constant supply voltage.

Anahtar Kavram

Self-Inductance and Inductive Reactance in AC Circuits

Tahmini Süre:1m 0s

Soru 175Soru →

An alternating current (AC) source with an RMS voltage of $200\ \text{V}$ is connected in series with a $16\ \Omega$ resistor, an inductor of inductive reactance $X_L = 18\ \Omega$ , and a capacitor of capacitive reactance $X_C = 30\ \Omega$ . What is the RMS current flowing in the circuit?

10.0\ \text{A}

3.13\ \text{A}

12.5\ \text{A}

16.67\ \text{A}

Cevabı ve açıklamayı göster

Cevap: $10.0\ \text{A}$

Cevap

The RMS current flowing in the circuit is

10.0\ \text{A}

.

To find the RMS current in a series RLC AC circuit, first calculate the net reactance:

|X_L - X_C| = |18\ \Omega - 30\ \Omega| = 12\ \Omega

. Next, determine total impedance using phasor addition:

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{16^2 + 12^2} = 20\ \Omega

. Finally, apply Ohm's law for AC:

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{200}{20} = 10.0\ \text{A}

.

Adım Adım Çözüm

1

Calculate the net reactance of the series circuit

|X_L - X_C| = |18\ \Omega - 30\ \Omega| = 12\ \Omega

Inductive and capacitive reactances are

180^\circ

out of phase in a series AC circuit.

2

Calculate the total impedance Z of the circuit

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{16^2 + (-12)^2} = \sqrt{256 + 144} = \sqrt{400} = 20\ \Omega

Impedance is the phasor sum of resistance and net reactance.

3

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

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{200\ \text{V}}{20\ \Omega} = 10.0\ \text{A}

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

Anahtar Kavram

Impedance and RMS Current in Series RLC AC Circuits

Soru 176Soru →

At a geographical survey location, the horizontal component of the Earth's magnetic field is measured as $3.0 \times 10^{-5}\text{ T}$ . If the angle of dip at this location is $60^\circ$ , what is the magnitude of the vertical component of the Earth's magnetic field?

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

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

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

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

Cevabı ve açıklamayı göster

Cevap: $5.2 \times 10^{-5}\text{ T}$

Cevap

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

The vertical component (

B_v

) and horizontal component (

B_h

) of Earth's magnetic field are related by the angle of dip (

\theta

) through

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

. Substituting

B_h = 3.0 \times 10^{-5}\text{ T}

and

\theta = 60^\circ

yields

B_v = 3.0 \times 10^{-5} \times 1.732 = 5.2 \times 10^{-5}\text{ T}

.

Adım Adım Çözüm

1

Identify the relationship between horizontal component (

B_h

), vertical component (

B_v

), and angle of dip (

\theta

).

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

The dip angle

\theta

represents the inclination of the total magnetic field relative to the horizontal plane.

2

Rearrange the equation to solve for the vertical component

B_v

.

B_v = B_h \times \tan\theta

Multiplying both sides by

B_h

isolates

B_v

.

3

Substitute the given values into the formula and calculate.

B_v = 3.0 \times 10^{-5}\text{ T} \times \tan 60^\circ = 3.0 \times 10^{-5}\text{ T} \times 1.732 = 5.196 \times 10^{-5}\text{ T} \approx 5.2 \times 10^{-5}\text{ T}

Using

\tan 60^\circ \approx 1.732

gives the vertical magnetic field magnitude.

Anahtar Kavram

Components of Earth's Magnetic Field and Angle of Dip

Tahmini Süre:1m 15s

Soru 177Soru →

At a certain research station, the vertical component of the Earth's magnetic field is measured as $2.4 \times 10^{-5}\text{ T}$ . If the angle of dip at this station is $37^\circ$ (given $\sin 37^\circ = 0.60$ and $\cos 37^\circ = 0.80$ ), calculate the horizontal component of the Earth's magnetic field in $\mu\text{T}$ .

Cevabı ve açıklamayı göster

Cevap: 32

Cevap

The horizontal component of the Earth's magnetic field is

32\ \mu\text{T}

.

The horizontal component

B_h

and vertical component

B_v

of the Earth's magnetic field are related by

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

, where

\theta

is the angle of dip. Given

B_v = 2.4 \times 10^{-5}\text{ T} = 24\ \mu\text{T}

and

\tan 37^\circ = 0.75

, rearranging gives

B_h = \frac{24}{0.75} = 32\ \mu\text{T}

.

Adım Adım Çözüm

1

Express the relationship between the vertical component (

B_v

), horizontal component (

B_h

), and angle of dip (

\theta

).

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

By definition of the angle of dip in the magnetic meridian, the tangent of the dip angle equals the ratio of the vertical component to the horizontal component.

2

Calculate

\tan 37^\circ

using the provided trigonometric values.

\tan 37^\circ = \frac{0.60}{0.80} = 0.75

Tangent of an angle is the ratio of sine to cosine of that angle.

3

Convert

B_v

from teslas to microteslas.

B_v = 2.4 \times 10^{-5}\text{ T} = 24\ \mu\text{T}

Since

1\ \mu\text{T} = 10^{-6}\text{ T}

,

2.4 \times 10^{-5}\text{ T} = 24 \times 10^{-6}\text{ T} = 24\ \mu\text{T}

.

4

Rearrange the equation to solve for

B_h

and substitute the values.

B_h = \frac{B_v}{\tan \theta} = \frac{24\ \mu\text{T}}{0.75} = 32\ \mu\text{T}

Dividing

24

by

0.75

yields

32

.

Anahtar Kavram

Resolution of Earth's magnetic field into horizontal and vertical components

Soru 178Soru →

An alternating current (AC) circuit consists of a resistor of resistance $R = 30\ \Omega$ , an inductor of inductive reactance $X_L = 80\ \Omega$ , and a capacitor of capacitive reactance $X_C = 40\ \Omega$ connected in series across an AC source of RMS voltage $150\ \text{V}$ . What is the average power dissipated in the circuit in watts?

Cevabı ve açıklamayı göster

Cevap: 270

Cevap

The average power dissipated in the circuit is

270\ \text{W}

.

The total impedance of a series RLC circuit is

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

. The RMS current is

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{150}{50} = 3\ \text{A}

. Because pure inductors and capacitors consume zero average power over a complete cycle, power is dissipated only across the resistor, giving

P = I_{\text{rms}}^2 R = 3^2 \times 30 = 270\ \text{W}

.

Adım Adım Çözüm

1

Calculate net reactance

X = 40\ \Omega

Inductive and capacitive reactances oppose each other in phase, so net reactance is

X_L - X_C

.

2

Calculate total circuit impedance

Z = 50\ \Omega

Resistance and net reactance add in quadrature:

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

.

3

Calculate RMS current

I_{\text{rms}} = 3\ \text{A}

Ohm's law for AC circuits gives

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

.

4

Calculate average power dissipated

P = 270\ \text{W}

Power is dissipated exclusively by resistance in an AC circuit:

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

.

Anahtar Kavram

Power Dissipation in AC Circuits

Tahmini Süre:1m 30s

Soru 179Soru →

At a geomagnetic observation post in West Africa, the total intensity of the Earth's magnetic field is $5.0 \times 10^{-5}\text{ T}$ and its vertical component is $4.0 \times 10^{-5}\text{ T}$ . What is the horizontal component of the Earth's magnetic field at this location?

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

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

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

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

Cevabı ve açıklamayı göster

Cevap: $3.0 \times 10^{-5}\text{ T}$

Cevap

The horizontal component of the Earth's magnetic field is

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

.

The Earth's total magnetic field

B

is resolved into perpendicular components: horizontal (

B_h

) and vertical (

B_v

). Applying the Pythagorean theorem

B^2 = B_h^2 + B_v^2

yields

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

.

Adım Adım Çözüm

1

Identify the relationship between total magnetic field intensity and its orthogonal components.

The total magnetic field

B

, horizontal component

B_h

, and vertical component

B_v

form a right triangle:

B^2 = B_h^2 + B_v^2

.

The horizontal and vertical components of the Earth's magnetic field are mutually perpendicular.

2

Rearrange the equation to solve for the horizontal component

B_h

.

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

Isolating

B_h

allows direct substitution of the given values.

3

Substitute the values

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

and

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

into the equation.

B_h = \sqrt{(5.0 \times 10^{-5})^2 - (4.0 \times 10^{-5})^2} = \sqrt{(25 - 16) \times 10^{-10}} = \sqrt{9 \times 10^{-10}} = 3.0 \times 10^{-5}\text{ T}

.

Evaluating the square root yields the exact magnitude of the horizontal magnetic field component.

Anahtar Kavram

Resolution of Earth's magnetic field into mutually perpendicular horizontal and vertical components using vector geometry.

Soru 180Soru →

When a straight metallic conductor of length $L$ moves at a constant velocity $v$ perpendicular to a uniform magnetic field $B$ , free electrons inside the conductor accumulate at one end, creating an internal electric field that eventually balances the magnetic force acting on them.

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Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

True. Moving a conductor through a magnetic field exerts a Lorentz magnetic force on its free electrons, pushing them toward one end. This separation of charge produces an internal electric field that exerts an opposing electrostatic force, reaching equilibrium when

qE = qvB

and resulting in an induced motional e.m.f. of

\mathcal{E} = BLv

.

The statement is correct because free charge carriers in a conductor moving through a magnetic field experience a magnetic Lorentz force. This force drives electrons to one side of the conductor, leaving positive ions on the other. The resulting charge separation builds an internal electric field

E

until the electric force

qE

equals the magnetic force

qvB

, creating a stable motional e.m.f.

\mathcal{E} = BLv

.

Adım Adım Çözüm

1

Analyze the force acting on free electrons due to motion in a magnetic field.

Each free electron carrying charge

q

experiences a magnetic force of magnitude

F_m = qvB

directed along the length of the conductor.

According to the Lorentz force law, a charge moving with velocity

v

perpendicular to a magnetic field

B

experiences a magnetic force perpendicular to both motion and magnetic field.

2

Determine the consequence of electron movement within the conductor.

Electrons accumulate at one end, making that end negatively charged and leaving the opposite end positively charged.

The conductor has finite boundaries, so mobile charge carriers migrate until stopped by the physical ends of the rod.

3

Evaluate the electric field and equilibrium condition established by charge separation.

An electric field

E

is formed pointing from the positive end to the negative end, creating an opposing electrostatic force

F_e = qE

. Accumulation stops when

F_e = F_m

, leading to

E = vB

and motional e.m.f.

\mathcal{E} = EL = BLv

.

Steady-state motional e.m.f. requires electrostatic equilibrium between the magnetic force driving charges apart and the electric force pulling them back.

Anahtar Kavram

Motional Electromotive Force and Microscopic Charge Separation

Electricity and Magnetism

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