AP Unit 12 Solutions

Outside the field leaves N and enters S; through the magnet it returns S to N. At the point just left of N, the compass north end points left, away from N. At the point just right of S, it also points left, toward S.

Both pieces remain dipoles; their cut faces carry opposite polarities. Two facing N ends repel. Moving farther weakens the dipole field.

A is ferromagnetic, such as iron: domains can align and retain magnetization. B is paramagnetic, such as aluminum: dipoles align weakly and do not retain alignment. C shows net diamagnetism: electronic response produces an opposite contribution.

All three have a diamagnetic contribution even if it is masked in A and B; permanent and induced magnetism are whole-system consequences of dipole alignment.

Permeability describes magnetization response to an applied field. Domain configurations, temperature, orientation, and saturation at high field can change that response.

μ₀ is the free-space constant appearing in AP field relationships; it is not the condition-dependent permeability of sample A.

Earth is approximately a magnetic dipole. Its region near geographic north acts approximately as magnetic south, aligning the north-seeking end toward it.

For positive charge, v × r is out of the page at U, into the page at D, and zero at F. U and D are perpendicular to v, so the source-field magnitude is greatest there.

Increasing speed strengthens B; increasing distance weakens B; changing charge sign reverses B.

The first right-moving positive object creates an out-of-page magnetic field at U. For the second positive object, index along its rightward velocity and middle along that out-of-page field give a downward force.

If the second object is stationary, its magnetic force is zero. An electric force from the first charge may still act.

$|q|vB=mv^2/r\quad\Rightarrow\quad r=mv/(|q|B)$. The radius uses |q| and is positive for either charge sign.

F = |q|vB = 3.20 × 10⁻¹⁵ N. The positive-charge direction is up, so the electron force is down.

r = mv/(|q|B) = 2.85 × 10⁻⁴ m. Speed stays constant because magnetic force is perpendicular to motion.

The magnetic force is down. The electron’s electric force must be up, so E points down. E = vB = 2.00 × 10⁴ V/m.

Parallel and antiparallel give zero force. An oblique velocity gives a nonzero force smaller than the original perpendicular maximum at the same speed.

B = μ₀I/(2πr) = 2.50 × 10⁻⁵ T. Upward current gives into-page B on the right. The field is tangent to a concentric circle, with neither radial nor axial component.

B versus r is an inverse curve. B versus 1/r is straight through the origin, with slope μ₀I/(2π) = 1.00 × 10⁻⁶ T·m.

Use a horizontal r axis from 0 to 0.0800 m and vertical B axis from 0 to at least 50.0 μT for the inverse curve. For the linear graph use horizontal 1/r from 0 to 50.0 m⁻¹ and the same B scale; the three points lie on B = (1.00 μT·m)(1/r). The slope is 1.00 μT·m = 1.00 × 10⁻⁶ T·m.

New B = (3/2)(25.0 μT) = 37.5 μT.

Curling right fingers with counterclockwise current gives out-of-page B at the center. The force question concerns the separately specified external rightward field.

Counterclockwise current is upward on the right side and downward on the left. Each vertical force has magnitude IℓB = 0.0360 N: right side into the page and left side out of the page.

Horizontal sides are parallel or antiparallel to B and feel zero force. The whole loop has zero net force, but the opposite vertical forces form a torque and can turn it.

Initially their tangent fields at the midpoint are equal and opposite, so net B is zero. Reversing one current reverses its field, so the midpoint fields add.

Flux = BA because the signed field is along the chosen normal. Values are +0.00800, +0.0200, +0.0200, and −0.00400 Wb.

For an open loop either normal may be chosen consistently. A closed surface uses an outward normal at each patch.

Positive circulation is counterclockwise for the out-of-page normal. Emfs from negative flux slopes are −0.0400 V, 0, and +0.0800 V.

Current is 0.0100 A clockwise, zero, and 0.0200 A counterclockwise. In the last interval the induced field is out of the page throughout, resisting the falling signed flux even after the external B changes sign.

Area changes at ΔA/Δt = ℓv. With B normal to the loop, |ℰ| = BΔA/Δt = Bℓv = 0.180 V.

I = 0.180/3.00 = 0.0600 A, up the rod and counterclockwise.

The magnetic force on the rod is leftward with magnitude IℓB = 0.00540 N, so the external force is 0.00540 N rightward.

P = Fv = 0.0108 W and I²R = 0.0108 W.

Positive carriers feel upward magnetic force, so the upper edge accumulates positive charge and is at higher potential. Balancing transverse forces gives E = vB, and |ΔV| = Ew = vBw.

B = |ΔV|/(vw) = 6.00 × 10⁻⁶/(12.0 × 0.0200) = 2.50 × 10⁻⁵ T. I = 2πrB/μ₀ = 5.00 A.

Twice r halves B, hence halves the Hall voltage to 3.00 μV.

Fix probe orientation tangent to the expected field, measure distance from wire axis, keep return lead far away, and work near the middle of a long straight section. Measure the same position with +I and −I and use [B(+I)−B(−I)]/2 to remove the stable background.

At fixed r take several current settings; graph isolated B versus I. At fixed I take several r settings; graph B versus 1/r. Repeat measurements and include zero-current checks.

Br values are 4.08 × 10⁻⁷, 3.92 × 10⁻⁷, and 4.08 × 10⁻⁷ T·m. The prediction μ₀I/(2π) is 4.00 × 10⁻⁷ T·m.

Probe position uncertainty is especially consequential at the smallest r; probe angular alignment and nearby current paths also matter. The data support the long-wire inverse-distance model within a few percent, provided uncertainties are comparable.

Current is counterclockwise, up the rod. The magnetic force on the rod points left; the external force at steady speed points right.

$\mathcal E=B\ell v,\quad I=B\ell v/R,\quad F=I\ell B=B^2\ell^2v/R$. Then $P=Fv=B^2\ell^2v^2/R$. Thus doubling v quadruples power, although emf, current, and force each double.

An open rod still has a motional potential difference but essentially no steady circuit current, so the induced IℓB drag and resistive heating vanish in the ideal model.

At the same constant speeds compare pulling force with the circuit open and closed; subtract the open-circuit baseline. Hold mechanical alignment fixed and reverse B as a check: induced drag still opposes motion.