Magnetism and Electromagnetism Lesson Test

Answer: B

Correct answer: They push apart. Every magnet has a north and a south pole, and the field lines from two facing north poles crowd against each other instead of joining, producing a repulsive force. Opposite poles, north facing south, connect their field lines and pull together. A handy memory hook is that magnets behave like electric charges: like repels, unlike attracts. Note that a magnet attracts plain iron with either pole, so only repulsion proves that two objects are both magnets.

• A: Magnets this close do exert a noticeable force on each other; the poles are not neutral toward one another.
• B (correct): Facing ends that are both north are like poles, and like poles repel.
• C: This guesses a sideways motion; the basic rule for facing like poles is simply a push apart.
• D: Pulling together happens with unlike poles, such as a north end facing a south end.

Answer: D

Correct answer: Two smaller magnets, each with a north and a south pole. A magnet's field comes from many tiny aligned regions, each with both poles. Cutting the bar leaves aligned regions in each piece, so each half becomes a complete magnet with its own north and south pole.

• A: This is the classic trap; an isolated single magnetic pole is not produced by cutting.
• B: Cutting does not scramble the aligned regions inside, so both pieces stay magnetized.
• C: A magnet always has opposite poles; a piece with two north poles and no south pole does not form.
• D (correct): Each piece still contains aligned magnetic regions, so each forms its own pair of poles.

Answer: D

Correct answer: Tesla. Field strength B is measured in teslas (T). Flux, which is B times area, is in webers, and inductance is in henries.

• A: The weber is the unit of magnetic flux, which is field times area, not field strength.
• B: Ampere-turns describe a coil's magnetizing strength, not the flux density it produces.
• C: The henry is the unit of inductance of a coil.
• D (correct): Magnetic flux density B is measured in teslas.

Answer: C

Correct answer: A steel washer. Strong attraction happens mainly with ferromagnetic materials such as iron, most steels, nickel and cobalt. Of these parts only the steel washer is ferromagnetic; copper, aluminum and brass are not.

• A: Brass is a copper-zinc alloy and is not attracted in any noticeable way.
• B: Aluminum conducts well but shows essentially no pull toward an ordinary magnet.
• C (correct): Steel is mostly iron, a ferromagnetic material whose domains the magnet can align.
• D: Copper is a good conductor but is not noticeably attracted to a magnet.

Answer: A

Correct answer: Like a magnetic south pole. Unlike poles attract. Since the compass needle's north end is drawn toward Earth's northern magnetic pole, that pole must behave like a magnetic south pole.

• A (correct): The needle's north end is attracted to it, and only an unlike pole attracts a north end.
• B: A north pole would repel the needle's north end, so this cannot be how it behaves.
• C: Magnetic poles are not electric charges; a compass does not respond to static charge.
• D: A region with no field would not steer the needle at all.

Answer: B

Correct answer: The field is stronger near that end. In a field diagram, the closer the lines, the stronger the field. Lines crowded near the end mean the field there is stronger than at the middle of the side.

• A: Field lines never cross, because a point cannot have two field directions.
• B (correct): Crowded lines show a stronger field and spread-out lines show a weaker one.
• C: This reverses the rule; closer lines mean a stronger field, not a weaker one.
• D: Line spacing shows strength, not direction; direction is given by the arrows.

Answer: B

Correct answer: Henry. When current through a coil changes, its changing magnetic field induces a voltage that pushes against the change; this self-induction is called inductance. One henry means a current change of one ampere per second induces one volt. The likely confusion is with resistance in ohms, which opposes steady current, or capacitance in farads, which stores charge. Think of an inductor as electrical inertia: it resists speeding up or slowing down the current, not the current itself.

• A: Farads measure capacitance, which stores energy in an electric field.
• B (correct): The property is inductance, and inductance is measured in henries.
• C: Ohms measure resistance, which opposes current itself rather than changes in current.
• D: Teslas measure magnetic flux density, not a coil's opposition to current change.

Answer: D

Correct answer: It makes the coil an electromagnet that pulls an armature to move the contacts. A relay is an electrically operated switch. Current in its coil turns the coil into an electromagnet, which pulls a hinged iron armature that opens or closes contacts in a completely separate circuit.

• A: That describes a thermal device, not a relay, which works by magnetic pull.
• B: That describes a transformer; a relay simply closes contacts on a load circuit that has its own supply.
• C: The load is on a separate circuit; the coil current never passes through it.
• D (correct): The coil's magnetic pull moves the armature, which opens or closes a separate circuit.

Answer: C

Correct answer: It stays dark. Induction requires a change in magnetic flux. With the magnet held still, the flux through the coil does not change, so no emf is induced and the lamp stays dark.

• A: This treats the magnet as a battery that runs down; a still magnet produces no current at all.
• B: Flickering would need a changing flux, but nothing in this setup is changing.
• C (correct): The flux through the coil is large but steady, and a steady flux induces no emf.
• D: A steady glow would need a continuous source of energy; a still magnet does not supply one.

Answer: C

Correct answer: As a generator, changing mechanical motion into electrical energy. Decide which way energy flows. Mechanical motion goes in through the crank and electrical energy comes out to light the lamp, so the machine is working as a generator.

• A: A transformer has no rotating shaft and takes electrical energy in, not mechanical work.
• B: A motor takes electrical energy in; here the energy enters as motion from the crank.
• C (correct): Mechanical work goes in through the crank and electrical energy comes out to the lamp.
• D: A relay only opens or closes contacts; it does not convert crank motion into electricity.

Answer: A

Correct answer: 90 ampere-turns. Magnetizing strength is $NI$.

The coil provides 90 ampere-turns.

• A (correct): Ampere-turns are turns times current: 150 times 0.6 equals 90.
• B: This misplaces the decimal in the other direction, as if the current were 6 A.
• C: This is a decimal-place slip in multiplying 150 by 0.6.
• D: This comes from dividing 150 by 0.6 instead of multiplying.

Answer: C

Correct answer: 100 microtesla. Convert 3 cm to 0.03 m. The $\pi$ cancels, leaving $2\times 10^{-7}$ times I divided by r.

That is 100 microtesla.

• A: This uses 0.003 m, moving the decimal one place too far when converting 3 cm.
• B: This drops the 2 in the denominator, using $\pi r$ instead of $2\pi r$.
• C (correct): Converting 3 cm to 0.03 m and substituting gives $1\times 10^{-4}$ T, which is 100 microtesla.
• D: This uses 3 m instead of 0.03 m, a centimeter-to-meter slip.

Answer: B

Correct answer: 20 microtesla. For a long wire, B is proportional to 1/r. Moving from 3 cm to 9 cm triples the distance, so the field becomes

microtesla.

• A: This multiplies by 3, treating the field as if it grew with distance.
• B (correct): The distance tripled, and the field is inversely proportional to distance, so it drops to one third.
• C: This halves the field, as if the distance had only doubled.
• D: This divides by 9, treating the field as inversely proportional to the square of the distance.

Answer: B

Correct answer: 0.5 N. The wire is perpendicular to the field, so

• A: This multiplies only B and I and leaves out the wire length.
• B (correct): Using $F=BI\ell$ with the wire perpendicular gives 0.4 times 5 times 0.25, or 0.5 N.
• C: This is a decimal slip in multiplying 0.4, 5 and 0.25.
• D: This divides by the length instead of multiplying by it.

Answer: C

Correct answer: 0.6 N. Use

The perpendicular value is

• A: This ignores the angle and uses the perpendicular value.
• B: This uses $\cos 30^{\circ }$ instead of $\sin 30^{\circ }$.
• C (correct): $F=BI\ell \sin \theta =0.5\times 6\times 0.4\times 0.5=0.6$ N.
• D: Zero force only occurs when the wire runs exactly along the field, at 0 degrees.

Answer: C

Correct answer: 6 × 10⁻⁴ N. The charge moves straight across the field, so

• A: This slips one power of ten in the multiplication.
• B: Only a charge at rest, or one moving along the field, feels no magnetic force; this one moves across it.
• C (correct): $F=qvB=3\times 10^{-6}\times 400\times 0.5=6\times 10^{-4}$ N.
• D: This divides by B instead of multiplying by it.

Answer: A

Correct answer: 0.03 Wb. With the field straight through the loop,

• A (correct): Flux is B times A when the field is along the loop's normal: 0.6 times 0.05 equals 0.03 Wb.
• B: This divides B by the area instead of multiplying.
• C: This is a decimal slip in multiplying 0.6 by 0.05.
• D: The number is right but the unit is not; tesla is field strength, while flux is in webers.

Answer: B

Correct answer: 0.02 Wb. Use

with $\theta$ measured from the normal.

• A: This ignores the tilt and treats the field as straight through the loop.
• B (correct): $\Phi =BA\cos 60^{\circ }=0.4\times 0.1\times 0.5=0.02$ Wb.
• C: This uses $\sin 60^{\circ }$ instead of $\cos 60^{\circ }$.
• D: Zero flux would need the field to lie in the loop's plane, at 90 degrees to the normal.

Answer: D

Correct answer: 8 V. Apply Faraday's law.

• A: This is a decimal slip; the rate of change is 0.04 Wb/s, not 0.4 Wb/s.
• B: This multiplies by the time instead of dividing by it.
• C: This finds the rate of flux change but forgets to multiply by the 200 turns.
• D (correct): $200\times 0.002/0.05=200\times 0.04=8$ V.

Answer: D

Correct answer: 36 V. Emf is proportional to $\Delta \Phi /\Delta t$. Cutting the time from 0.3 s to 0.1 s makes the rate three times larger, so the emf becomes

• A: This divides by three, as if a faster change made a smaller emf.
• B: This doubles the emf, but the time was cut to one third, not one half.
• C: This ignores the rate; Faraday's law depends on how fast the flux changes.
• D (correct): The change happens three times faster, so the emf is three times larger.

Answer: C

Correct answer: 20 V. Use

• A: Zero voltage applies only to a steady current; here the current is changing.
• B: This divides the rate by the inductance instead of multiplying.
• C (correct): $V=L\Delta I/\Delta t=0.5\times 40=20$ V.
• D: This divides the inductance by the rate instead of multiplying.

Answer: D

Correct answer: 0.9 J. Use

• A: This forgets to square the current.
• B: This leaves out the factor of one half.
• C: This simply multiplies L by I, missing both the half and the square.
• D (correct): $W=\frac{1}{2}L{I}^2=0.5\times 0.2\times 9=0.9$ J.

Answer: C

Correct answer: 0.03 s. The RL time constant is

• A: This multiplies L by R instead of dividing.
• B: This divides R by L, which inverts the formula.
• C (correct): $\tau =L/R=0.45/15=0.03$ s.
• D: This is five time constants, about how long the current takes to settle fully.

Answer: A

Correct answer: Up the page. A cross means current into the page. With the right thumb pointing into the page, the fingers curl clockwise. Following a clockwise circle, the left side moves upward, so the needle's north end points up the page.

• A (correct): A cross means current into the page, giving a clockwise field; on the left side of a clockwise circle the direction is upward.
• B: The field circles the wire; it does not point straight out from it.
• C: This would be correct for a dot, which gives a counterclockwise field.
• D: Field lines around a wire are circles, so the needle points along the circle, not toward the wire.

Answer: B

Correct answer: Reluctance. Reluctance opposes flux the way resistance opposes current. Iron has low reluctance and air has high reluctance, so adding an air gap raises the reluctance of the path.

• A: Hysteresis is the lag of magnetization behind the driving field, not the opposition of the path.
• B (correct): Reluctance is the opposition a magnetic path offers to flux, and an air gap raises it sharply.
• C: Ampere-turns depend on the coil's turns and current, which the gap does not change.
• D: Air has far lower permeability than iron, so the gap lowers the path's ability to carry flux.

Answer: D

Correct answer: Its domains align easily and also relax when the current stops. A soft iron core greatly boosts the coil's field because its domains line up easily. Those domains also relax when the current stops, so the electromagnet turns off cleanly, which is what a switchable lock needs.

• A: Keeping magnetism is a property of hard steel, and it would stop the lock from releasing cleanly.
• B: The core's job is magnetic, not to carry current; conductivity is not the reason for the choice.
• C: The core does not supply energy; the coil current must still flow.
• D (correct): Soft iron strengthens the field while on and loses most of its magnetism when switched off.

Answer: D

Correct answer: Clockwise. By Lenz's law, the induced current opposes the change. The into-page flux is decreasing, so the induced field points into the page to support it. By the grip rule, a field into the page inside a loop needs clockwise current.

• A: The flux falls steadily in one direction, so the induced current keeps one direction.
• B: Counterclockwise current makes a field out of the page, which would speed up the loss instead of opposing it.
• C: The flux is changing, so an emf is induced and current flows in the closed loop.
• D (correct): The induced field must point into the page to prop up the falling flux, and that requires clockwise current.

Answer: B

Correct answer: It swings left. An emf appears only while the flux changes. Pushing in increases the flux and pulling out decreases it, so the induced current, and the needle swing, is in the opposite direction: to the left.

• A: The magnet is moving, so the flux is changing and an emf is induced.
• B (correct): Pulling out decreases the flux, the opposite change from pushing in, so the induced current reverses.
• C: Speed affects the size of the swing, not its direction; the change is now the opposite one.
• D: Swinging the same way would need the flux to increase again, but pulling out decreases it.

Answer: A

Correct answer: About zero, because the core flux is no longer changing. A transformer works by induction, which needs changing flux. A steady DC current creates a constant flux in the core, so after the instant of switching on, no voltage is induced in the output coil.

• A (correct): Steady DC makes a constant flux, and a constant flux induces no voltage in the output coil.
• B: Nothing in the circuit is alternating, so no AC voltage can appear.
• C: Sharing a core does not induce voltage by itself; the flux must be changing.
• D: A transformer cannot pass steady DC, since only changing flux induces voltage.

Answer: C

Correct answer: To block eddy currents that waste energy as heat. Changing flux induces eddy currents in a solid core, and those currents waste energy as heat. Thin insulated sheets, called laminations, block those current loops and reduce the loss.

• A: Splitting the core into sheets does not raise how much flux the iron can carry before it saturates.
• B: The coils are linked magnetically; the core is not meant to carry current between them.
• C (correct): The insulated layers break up the swirling current loops that changing flux drives in solid metal.
• D: Laminations are about reducing induced currents, not about holding permanent magnetism.

Answer: A

Correct answer: 3 A. Coil X:

ampere-turns. Twice that is 240 ampere-turns.

• A (correct): Coil X gives 120 ampere-turns, so twice that is 240 ampere-turns, and 240 divided by 80 turns is 3 A.
• B: This doubles the target twice, aiming for four times coil X's strength.
• C: This scales the current down with the turns instead of up.
• D: This only matches coil X's strength and forgets to double it.

Answer: D

Correct answer: 24 microtesla. B is proportional to I/r. The current ratio is

and the distance ratio is

• A: This multiplies by 1.5 for both changes, treating a larger distance as a stronger field.
• B: This applies the larger distance but forgets the larger current.
• C: This applies the current increase but forgets that the sensor moved farther away.
• D (correct): Current rises by a factor of 1.5 and distance by a factor of 1.5, so the two effects cancel.

Answer: D

Correct answer: 8 V. The change in field is

T, so

Wb per turn.

• A: This finds the rate of flux change per turn but forgets to multiply by 40 turns.
• B: This uses the final field of 0.1 T as the change instead of the 0.4 T drop.
• C: This uses the starting field of 0.5 T as the change instead of the 0.4 T drop.
• D (correct): The flux change per turn is 0.4 times 0.03, or 0.012 Wb, and 40 times 0.012 divided by 0.06 is 8 V.

Answer: A

Correct answer: 13.5 J. Since

energy scales with the square of the current. The current goes from 2 A to 6 A, a factor of 3, so the energy grows by

• A (correct): The current triples, and energy goes with current squared, so the energy grows by 9: 1.5 times 9 is 13.5 J.
• B: This scales by the cube of the ratio, 27, instead of the square.
• C: This multiplies 1.5 by the new current of 6 instead of scaling by the square of the ratio.
• D: This triples the energy, forgetting that the current is squared.

Answer: A

Correct answer: 0.04 s and about 0.76 A. The time constant is

The final current is

and after one time constant the current reaches about 63 percent of it.

• A (correct): $\tau =0.8/20=0.04$ s; the final current is 1.2 A, and about 63 percent of that is 0.76 A.
• B: This multiplies L by R instead of dividing to get the time constant.
• C: 0.2 s is about five time constants, the time to settle fully, not one time constant.
• D: The time constant is right, but 1.2 A is the final current, which is not reached after one time constant.

Answer: B

Correct answer: Into the page. The force must be perpendicular to both the field and the current, so it is either into or out of the page. Using the right-hand palm rule, point the fingers to the right along the field and the thumb up along the current; the palm, and the force, faces into the page.

• A: The force is never along the field; it is perpendicular to both field and current.
• B (correct): With fingers along the field and thumb along the current, the right palm faces into the page.
• C: The force is perpendicular to the current, so it cannot point along the wire.
• D: This is the direction the palm would face if the current or field were reversed.

Answer: B

Correct answer: The crank becomes harder to turn. By Lenz's law, the induced current produces a field that opposes the motion causing it. Drawing more current makes that opposition stronger, so the person must do more work, which matches the greater electrical energy delivered to two lamps.

• A: The opposing effect depends on the current drawn, not only on the magnets.
• B (correct): The larger induced current makes a stronger field that opposes the motion, so more work is needed.
• C: Easier turning while supplying more energy would create energy from nothing.
• D: A generator cannot drive itself; that would be energy from nothing.

Answer: A

Correct answer: It increases only slightly. Once nearly all domains in the core are aligned, the core is saturated. Further increases in coil current add only a small amount of extra flux, about what an air core would give, so doubling the current raises the flux only slightly.

• A (correct): With nearly all domains aligned, extra current adds only about what an air core would add.
• B: Saturation limits further gains; it does not wipe out the flux already present.
• C: This wrongly applies a squared relationship; flux does not grow with the square of current.
• D: Doubling would happen only well below saturation, where the iron can still respond fully.

Answer: A

A coil stores energy in its magnetic field. When the transistor turns off, the coil resists the sudden drop in current and produces a spike many times the supply voltage. A diode across the coil, pointed so it does not conduct normally, lets that current circulate and die away, clamping the spike.

• A (correct): The collapsing field produces a large spike, and a reverse-connected diode gives the coil current a safe path to die away.
• B: The control and load sides are isolated, and a higher coil voltage would make the spike worse.
• C: The failure happens at turn-off, from the inductive spike, and a series diode does not give the current a path to decay.
• D: A diode that conducts during normal operation would short the coil so the relay never pulls in.

Answer: C

Correct answer: The magnet is repelled, because the top face of the loop acts as a north pole. Curl the right-hand fingers counterclockwise, the way the current flows as seen from above; the thumb points up, so the top face of the loop is a north pole. Like poles repel, so the magnet's north end is pushed away.

• A: A current loop has poles like any magnet; whether it attracts or repels depends on which pole faces it.
• B: A current loop makes its own magnetic field and acts like a short magnet even without iron.
• C (correct): Curling right-hand fingers counterclockwise puts the thumb upward, so the top face is north, and north repels north.
• D: This reverses the coil rule; counterclockwise current seen from above makes the top face north.

Answer: B

Correct answer: Weber. Flux is

so its unit is teslas times square meters, which is named the weber (Wb). Keep tesla for field strength and weber for field through an area.

• A: The tesla measures flux density B, the field strength at a point, not the total field through an area.
• B (correct): Flux is field times area, and one weber equals one tesla multiplied by one square meter.
• C: The henry is the unit of inductance, a coil's opposition to changes in its current.
• D: Ampere-turns measure a coil's magnetizing strength, not the flux passing through a surface.

Answer: D

Correct answer: To the right. A dot means current comes out of the page, so with your right thumb pointing at you, your fingers curl counterclockwise. Tracing a counterclockwise circle, the path at the bottom moves from left to right, so the needle points to the right.

• A: This is the direction at the bottom of a clockwise circle, which belongs to a cross (current into the page).
• B: The field around a straight wire circles it; a compass beside the wire never points straight at it.
• C: Field lines around a wire are circles, so the needle lies along the circle, not outward from the wire.
• D (correct): A dot gives a counterclockwise field, and at the bottom of a counterclockwise circle the field runs to the right.

Answer: C

Correct answer: 5 × 10⁻⁴ N. Use

• A: This ignores the angle and treats the particle as moving straight across the field, where $\sin \theta =1$.
• B: This uses $\cos 30^{\circ }$; the magnetic force depends on the sine of the angle between velocity and field.
• C (correct): $qvB\sin 30^{\circ }=5\times 10^{-6}\times 800\times 0.25\times 0.5=5\times 10^{-4}$ N.
• D: This is 100 times too large, an exponent slip when multiplying the powers of ten.

Answer: D

Correct answer: 4 V. Turning the coil changes the angle, so the flux changes. At the start

Wb per turn; edge-on the flux is 0, so

Wb.

• A: This is the emf of a single turn; the change must be multiplied by all 25 turns.
• B: This multiplies by the time instead of dividing by it; a faster change gives a larger emf, so time goes in the denominator.
• C: Ending at zero flux does not mean zero emf; the emf depends on how much the flux changed, which here is the full 0.016 Wb per turn.
• D (correct): Flux per turn falls from 0.016 Wb to 0, and $25\times 0.016/0.1=4$ V.

Answer: C

Correct answer: Reversing the direction of the coil current. Wrap your right-hand fingers in the direction of the conventional current and your thumb points to the north end. If the current flows the other way around the coil, your thumb points the other way, so the poles flip. Turns, current size and core material only change strength.

• A: More turns raise the ampere-turns and make the magnet stronger, but the field direction stays the same.
• B: More current strengthens the field without changing which end is north.
• C (correct): By the right-hand coil rule, reversing the current reverses the field, so the ends trade poles.
• D: A different core changes how strong the magnet is and how much magnetism it keeps, not which end is north.

Answer: A

Correct answer: 4.5 J. Find the energy at each current with

then subtract.

The increase is

• A (correct): Final energy 4.8 J minus starting energy 0.3 J gives an increase of 4.5 J.
• B: This squares the change in current, 3 A; the energy formula must be applied to each current and then subtracted.
• C: This is the total energy stored at 4 A; the coil already held 0.3 J at 1 A.
• D: This uses the current change without squaring it; inductor energy grows with the square of the current.

Answer: B

Correct answer: About 1 s, settling at 3 A. The time constant is

and the current is essentially complete after about five time constants,

Once steady, the coil acts like a plain resistance, so

• A: 0.2 s is only one time constant, when the current has reached about 63 percent of its final value.
• B (correct): $\tau =2.4/12=0.2$ s, about five time constants is 1 s, and the final current is 36/12 = 3 A.
• C: 1.9 A is about 63 percent of the final value, the current after one time constant, not the steady current.
• D: This inverts the time constant to R/L = 5 and multiplies by five; the time constant is L divided by R.

Answer: A

Correct answer: 2 A. With the wire straight across the field,

Solve for current.

• A (correct): $I=F/(B\ell )=0.36/(0.3\times 0.6)=0.36/0.18=2$ A.
• B: This inverts the relationship, dividing $B\ell$ by the force instead of the force by $B\ell$.
• C: This multiplies force, field and length together instead of solving $F=BI\ell$ for I.
• D: This divides by the field but multiplies by the length; both B and $\ell$ belong in the denominator.

Answer: A

Correct answer: Permeability. Permeability measures how readily a material carries magnetic flux. A high-permeability core such as soft iron greatly multiplies the flux a coil produces, which is the same as giving the flux path low reluctance.

• A (correct): Permeability describes how easily a material supports magnetic flux; soft iron's is hundreds to thousands of times that of air.
• B: Reluctance is opposition to flux, so a good core path should have low reluctance, not high.
• C: Hysteresis is the lag of magnetization behind the field; more of it means more energy lost each AC cycle.
• D: Saturation is the limit where nearly all domains are aligned and extra current adds little flux; it is a limit, not a property to maximize.

Answer: D

Correct answer: 0 Wb. Use

where $\theta$ is measured from the line perpendicular to the loop. Field lines parallel to the surface are at 90 degrees to that line, so

Wb. No field passes through the loop.

• A: This is the flux only when the field passes straight through the loop face; here it runs along the surface.
• B: Besides ignoring the angle, this gives the answer in teslas, the unit of field strength, not of flux.
• C: This divides the field by the area instead of multiplying, and it also ignores the loop's orientation.
• D (correct): The angle between the field and the line perpendicular to the loop is 90 degrees, and $\cos 90^{\circ }=0$.