Magnetism and Electromagnetism

1. Outside a magnet, a compass's north-seeking end follows the field direction from north to south. Pointing away from the right end identifies that end as north.
2. The measured gradient is 1.0×10⁻³ T A⁻¹ = μ₀n, so n = 0.001/(4π×10⁻⁷)≈796 m⁻¹.
3. Reversing current reverses field direction. At −1.5 A the linear calibration predicts −1.5 mT; its magnitude is unchanged.
4. Control current rather than supply voltage, because resistance and heating may change. Keep probe orientation and centre position fixed, repeat measurements and subtract the ambient field.
5. Measure a current-up and current-down sequence to detect hysteresis or remanence in the iron core; the air-core proportional model need not describe a saturating core.
6. Field lines represent local directions and relative density. They are not physical wires or trajectories that isolated magnetic poles follow; a complete magnetic field map forms loops.

Answer (a) Right end is north; n≈796 turns m⁻¹; reversed field is −1.5 mT. (b) Vary core material at fixed coil geometry and current; measure signed field at fixed locations and check for residual magnetism.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. The torque factor is NBIA = 20(0.20)(2.0)(0.010) = 0.080 N m.
2. At θ = 30°, multiply by sin30° = 0.50 to obtain 0.040 N m. Torque is greatest when the normal is perpendicular to the field.
3. Instantaneous mechanical power is τω = 0.060(50) = 3.0 W. Opposite forces on opposite sides form a couple, even if the net force on the whole coil is zero.
4. Without reversal, the magnetic couple would reverse after half a turn and oppose continued rotation. Inertia carries the coil through positions of zero torque.
5. Electrical input is divided between mechanical output and losses such as I²R heating, bearing friction and magnetic losses; power output cannot exceed input.
6. Increasing current increases torque but raises resistive heating quadratically. A lifting motor needs sufficient torque while meeting temperature and efficiency constraints.

Answer (a) Maximum torque = 0.080 N m; at 30° = 0.040 N m; mechanical power = 3.0 W. (b) A split-ring commutator reverses coil current each half-turn, maintaining torque in the rotation direction.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. With the field normal to the coil, flux per turn is Φ = BA and flux linkage is NBA.
2. The initial field gradient is +2.0 T s⁻¹, so ε = −100(0.010)(2.0) = −2.0 V and I = ε/R = −0.40 A.
3. Constant field gives zero flux-change emf. The final gradient is −2.0 T s⁻¹, reversing both emf and current to +2.0 V and +0.40 A.
4. During each 0.20 s ramp, I²Rt = (0.40)²(5.0)(0.20) = 0.16 J; total heating is 0.32 J.
5. When the applied field decreases, the induced current creates a field in the original positive direction, opposing the decrease rather than always opposing the existing field.
6. Lenz's law gives a reaction opposing the flux change. The apparatus maintaining that change must supply energy that is dissipated in the coil; induction is not a free source of energy.

Answer (a) 0–0.2 s: ε = −2.0 V, I = −0.40 A; 0.2–0.5 s: zero; 0.5–0.7 s: ε = +2.0 V, I = +0.40 A. (b) Total heating = 0.32 J; the external apparatus changing the field supplies the energy.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. Vs/Vp = Ns/Np = 10000/250 = 40. At the sending end Is = P/Vs = 20000/10000 = 2.0 A.
2. For an ideal transformer input power equals output power, so Ip = 20000/250 = 80 A.
3. Line heating is I²R = 2²(5) = 20 W, leaving 20000−20 = 19980 W at the receiving end; the line voltage drop is IR = 10 V.
4. Using sending current P/V = 20000/250 = 80 A would give a 400 V line drop and I²R = 32000 W, exceeding the sending power.
5. This contradicts a passive receiving load at the assumed operating point; the calculation demonstrates why voltage drop and the load must be solved together, and why high-voltage transmission is useful.
6. A steady DC current produces steady flux after the switching transient, so it induces no sustained secondary emf. Real AC transformers also have core and winding losses and insulation limits.

Answer (a) Turns ratio = 40; secondary current = 2.0 A; primary current = 80 A; line loss = 20 W; received power = 19.98 kW. (b) The naive 250 V case demands 32 kW of line loss for only 20 kW sent, revealing that the assumed delivered operating point is impossible with a passive load.

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1. In the selector, qE = qvB gives v = E/B = 2.0×10⁵ m s⁻¹; charge cancels, so this stage selects speed rather than mass.
2. In the analyser qvB = mv²/r, so m = qBr/v = (1.60×10⁻¹⁹)(0.20)(0.0375)/(2.0×10⁵) = 6.0×10⁻²⁷ kg.
3. The magnetic force is perpendicular to velocity, changing its direction and providing the circular resultant without doing work on the ion.
4. The selected speed E/B does not depend on charge magnitude when both selector forces scale with q.
5. In the analyser r = mv/(|q|B), so doubling charge halves radius at fixed mass, speed and field. The apparatus measures mass-to-charge ratio, not mass independently.
6. Changing charge sign reverses the magnetic force direction. Field direction and incoming velocity must be specified to name the clockwise or counterclockwise sense; radius alone gives no sign.

Answer (a) Selected speed = 2.0×10⁵ m s⁻¹; ion mass = 6.0×10⁻²⁷ kg. (b) Its radius halves to 0.01875 m; an opposite-sign charge curves in the opposite direction.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.