Electromagnetic Effects Worked ExamplesIGCSE Physics 0625

(a) The field lines are concentric circles centred on the wire, in the plane of the card. They are closer together near the wire and further apart further away.

(b) Use the right-hand grip rule: point the right thumb in the direction of the current (towards you, out of the page); the fingers curl in the direction of the field. Seen from above, the field is anticlockwise.

(c) The field is stronger at P, because P is closer to the wire. The field strength decreases as the distance from the wire increases (the circles are further apart).

(d)(i) Reversing the current reverses the direction of the field (it becomes clockwise from above); the pattern of circles is unchanged. (ii) Increasing the current makes the field stronger at every point (field lines closer together); the pattern and direction are unchanged.

(a)(i) While the magnet moves into the coil, the needle deflects to one side (an e.m.f. is induced and a current flows).

(ii) While the magnet is held still, the reading is zero — the needle returns to the centre.

(iii) While the magnet is pulled out, the needle deflects the opposite way.

(b) The size of the induced e.m.f. depends on how fast the magnetic field linking the coil changes. Apart from speed: use a stronger magnet, or use a coil with more turns.

(c) An e.m.f. is induced only when the magnetic field linking the coil is changing (the wire cuts field lines). A stationary magnet gives a constant field through the coil, so no e.m.f. is induced.

(a) End X becomes a north pole.

(b) The direction of an induced e.m.f. opposes the change causing it. The change is the magnet moving away, so the induced current makes a pole at X that attracts the magnet, pulling back against its motion. A north pole attracts the magnet's south pole.

(If X became a south pole, it would repel the magnet and help it move away — the induced current would then be producing energy for nothing, which is impossible.)

(c) The meter deflects the opposite way. Now the change is the south pole approaching, so X becomes a south pole to repel it; the induced current is in the opposite direction.

The current is induced, so this is a generator situation: use Fleming's right-hand rule. Thumb = motion, first finger = field, second finger = induced current.

The wire must cut the field lines, so the motion must be perpendicular to the field. The field runs from N to S (left to right), so moving towards either pole (C or D) cuts no field lines and induces no current.

Right hand: first finger to the right (field), second finger out of the page (current). The thumb then points towards the bottom of the page.

Option A is the answer you get by wrongly using the left-hand (motor) rule, which gives the direction of a force, not of the motion that induces a current.

(a) Use Fleming's left-hand rule (force on a current in a field). First finger = field: into the page. Second finger = current: up the page along the rod.

The thumb, at right angles to both, points to the left — so the rod rolls to the left along the rails, away from the cell. The field, current and force are mutually perpendicular.

(b) Reversing the cell reverses the current, so the force reverses: the rod now moves to the right.

(c) Reversing the current reverses the force, and reversing the field reverses it again. Two reversals cancel, so the rod moves to the left, as in (a).

(d) Increase the current (for example a cell of higher e.m.f.) or use a stronger magnet.

(a) X is the split-ring commutator; Y is a brush (a carbon block).

(b)(i) The split-ring commutator reverses the direction of the current in the coil every half-turn (as the coil passes the position where its plane is perpendicular to the field). This keeps the turning effect in the same direction, so the coil keeps rotating the same way.

(ii) The brushes make a sliding electrical contact between the fixed circuit (the cell) and the rotating commutator, so current can reach the coil while it turns.

(iii) The coil and commutator are fixed to the axle, which allows the coil to rotate about it (and transfers the rotation to whatever the motor drives).

(c) The turning effect comes from the forces on the two sides of the coil, which act in opposite directions. It is increased by: more turns on the coil; a larger current; a stronger magnetic field.

(a) The rings are slip rings; the carbon blocks are brushes. Each slip ring stays connected to the same end of the coil, so the output alternates.

(b) As the coil rotates, its sides cut the magnetic field lines (the field linking the coil keeps changing), so an e.m.f. is induced.

(c) Any two of: rotate the coil faster; use a stronger magnet; put more turns on the coil.

(d) The soft-iron core strengthens the magnetic field through the coil, so the induced e.m.f. (the output) is larger.

(a) The alternating current in the primary coil produces a changing (alternating) magnetic field. The soft-iron core carries this changing field to the secondary coil. The secondary coil is therefore in a changing magnetic field, so an (alternating) e.m.f. is induced in it.

(b) Write the turns-ratio equation, then rearrange for $V_p$.

Substitute.

(c) For a 100% efficient transformer the power in equals the power out. Use the primary voltage for the primary current.

Substitute.

The turns ratio gives the voltage ratio.

This rules out C and D. For an ideal transformer, power in = power out, so if the voltage goes up by a factor of $8$, the current must go down by the same factor.

A would mean the output power is $64$ times the input power — impossible. The answer is B.

(a) Electrons are negative, so the conventional current is in the opposite direction to their motion: to the left.

The force on the beam is upwards (the direction of the deflection). Use Fleming's left-hand rule: second finger (current) to the left, thumb (force) upwards. The first finger (field) then points out of the page.

Check: the field must be at right angles to both the motion and the force, so it can only be into or out of the page — never up or down the page.

(b) Alpha particles are positively charged, so their conventional current is in their direction of motion (to the right) — opposite to that of the electrons. The force is reversed: they are deflected downwards.

(c) Gamma rays are uncharged (electromagnetic radiation), so there is no force on them: they pass straight through undeflected.

(a) The e.m.f. is largest (peak or trough) at Q and S. There the plane of the coil is parallel to the field, so its sides are moving at right angles to the field lines and cut them at the fastest rate. (At P, R and T the coil is perpendicular to the field; its sides move along the field lines, cutting none, so the e.m.f. is zero.)

(b) One complete cycle (P to T) is one revolution. Read its period from the graph, then use $f = \dfrac{1}{T}$.

One cycle per revolution, so the coil makes $25$ revolutions per second.

(c) Twice as many turns doubles the e.m.f.: the peaks become $+12\text{ V}$ and $-12\text{ V}$. The speed is unchanged, so the period stays $0.040\text{ s}$ and the frequency stays $25\text{ Hz}$.

(d) Rotating twice as fast makes the coil cut field lines twice as fast, so the peak e.m.f. doubles to $12\text{ V}$ and the period halves to $0.020\text{ s}$ (frequency $50\text{ Hz}$).

(a) Write the power-loss equation and make $I^2$ the subject (convert kW to W).

Take the square root — $225$ is $I^2$, not the current.

(b) Use $P = IV$ with the power transmitted, $3.3\text{ MW} = 3.3 \times 10^6\text{ W}$.

(c) First find the current at $11\text{ kV}$.

Then the power lost in the same $20\,\Omega$ line.

(d) For the same power transmitted, a higher voltage means a smaller current ($P = IV$). The power lost in the cables is $I^2R$, so a smaller current gives a much smaller loss: here $20$ times less current gives $400$ times less loss ($4.5\text{ kW}$ instead of $1.8\text{ MW}$ — more than half the power). The resistance of the line is the same in both cases.