⚡ بنك الأسئلة: الحث الكهرومغناطيسي (12A)

📌 \(\phi=BA\cos\theta\) where \(\theta\) is the angle between the field and the normal; the angle between the field and the surface is \(60^{\circ}\) so \(\theta=30^{\circ}\).

\(\phi=0.3\times0.2\times\cos30^{\circ}=0.052\,T\cdot m^{2}\) (or \(BA\sin60^{\circ}\)).

\(\phi_A=BA\cos20^{\circ}=0.5\Rightarrow BA=\dfrac{0.5}{0.940}=0.532\).

📌 When the field is perpendicular to the surface (parallel to the normal) \(\cos0=1\) so \(\phi_B=BA=0.53\,T\cdot m^{2}\), the maximum value.

📌 Lenz's law: the induced field opposes the change of flux. A magnet approaching with its north pole downward gives an increasing downward flux so an upward field is induced (counterclockwise); a magnet moving away with its north pole up gives a decreasing upward flux so an upward field is induced as well (counterclockwise).

📌 The wire's field at the loop points into the page; if the loop moves away the inward flux decreases so an inward field is induced, i.e. a clockwise current, and if it comes closer the opposite happens. So the answer is moving the loop away from the wire.

📌 The field inside the coil points to the left (its left end is a north pole), and the magnet approaches with its north pole, so its field at the coil points to the right, opposite to the coil's field.

📌 The total flux to the left decreases, so an induced emf opposes the decrease, i.e. it supports the original current, and the current increases at that moment.

📌 The required direction of the induced current in the second coil occurs when the flux in the first coil increases (for the winding in the figure), so we look for the case in which the flux increases.

📌 Opening the switch, removing the core or increasing the resistance reduces the flux (smaller current or lower permeability), whereas decreasing the resistance increases the current and the flux and gives the required direction.

📌 A clockwise current produces a field into the page inside the loop, opposing an increase of the outward flux.

📌 The outward flux increases when the area increases (expanding the loop); moving inside a uniform field or rotating while parallel to the page does not change it, and removing the loop decreases it, giving a counterclockwise current.

\(A=\pi r^{2}=\pi(0.2)^{2}=0.1257\,m^{2}\) and \(\dfrac{dB}{dt}=0.9t^{2}+2=0.9(9)+2=10.1\,T/s\).

\(\Delta V_{ind}=-A\dfrac{dB}{dt}=-0.1257\times10.1=-1.27\,V\).

\(|\Delta V_{ind}|=B\dfrac{\Delta A}{\Delta t}=0.5\times10^{-4}\times\dfrac{0.1}{0.4}=1.25\times10^{-5}\,V\).

📌 \(\phi=BA\cos\omega t\) so \(\Delta V_{ind}=-\dfrac{d\phi}{dt}=BA\omega\sin\omega t\).

\(=0.2\times0.1\times5\times\sin(5\times4)=0.1\times\sin(20\,rad)=0.1\times0.913\approx0.09\,V\) (the angle is in radians).

\(B=\dfrac{\mu_0NI}{\ell}=\dfrac{4\pi\times10^{-7}\times50\times3}{0.1}=1.885\times10^{-3}\,T\).

\(\phi=BA=1.885\times10^{-3}\times0.2=3.77\times10^{-4}\,T\cdot m^{2}\).

📌 Reversing the current means it changes from \(+2\,A\) to \(-2\,A\): \(\Delta I=-4\,A\).

\(\Delta V_{ind}=-L\dfrac{\Delta I}{\Delta t}=-0.1\times\dfrac{-4}{0.15}=+2.7\,V\).

📌 A blade is a rod rotating about one end: a point at distance \(x\) from the hub has speed \(v=\omega x\), so the emf is \(\int B\omega x\,dx=\dfrac{1}{2}B\omega L^{2}\).

\(\Delta V_{ind}=\dfrac{1}{2}\times0.4\times10^{-4}\times200\times4^{2}=0.064\,V\).

📌 From the graph: \(\phi_{max}=5\times10^{-2}\,T\cdot m^{2}\) and the flux falls from its maximum to zero in \(2\,s\) (a quarter cycle), so \(T=8\,s\) and \(\omega=\dfrac{2\pi}{8}=0.785\,rad/s\).

\(\Delta V_{max}=N\phi_{max}\omega=20\times5\times10^{-2}\times0.785=0.79\,V\).

\(N_s=\dfrac{30}{2}=15\) and the emf in the whole secondary is \(V_s=M\dfrac{\Delta I}{\Delta t}=0.2\times\dfrac{3}{0.1}=6\,V\).

\(V_s=N_s\dfrac{\Delta\phi}{\Delta t}\Rightarrow\dfrac{\Delta\phi}{\Delta t}=\dfrac{6}{15}=0.4\,V\).

\(\Delta V_{ind}=BLv=0.2\times0.2\times20=0.8\,V\).

\(I=\dfrac{\Delta V_{ind}}{R}=\dfrac{0.8}{5}=0.16\,A\).

📌 From the graph: \(\Delta V_{max}=2\,V\) at \(2\,s\) and it returns to zero at \(4\,s\) so a quarter cycle is \(2\,s\), hence \(T=8\,s\) and \(\omega=0.785\,rad/s\).

\(\phi_{max}=\dfrac{\Delta V_{max}}{N\omega}=\dfrac{2}{30\times0.785}=0.085\,T\cdot m^{2}\).

📌 The emf starts from zero and grows positive so \(\phi=\phi_{max}\cos\omega t\); at \(t=4\,s\), \(\omega t=\pi\) and \(\phi=-0.085\,T\cdot m^{2}\).

📌 Wood is an insulator so no induced current flows in it and the magnet falls freely.

📌 In the aluminium ring (a conductor) an induced current produces a field that opposes the motion of the magnet (Lenz's law), slowing it down, so the magnet above the wood arrives first.

📌 The flux reached when the magnet gets to the same position does not depend on the speed, so the flux through the coil does not increase because the speed is doubled (it only changes faster).

📌 But \(\Delta V_{ind}=-N\dfrac{\Delta\phi}{\Delta t}\) doubles since \(\Delta t\) halves, so the current and the deflection increase.

📌 When the switch is closed the current begins to grow in the coil branch and an induced emf opposes its growth: \(\Delta V_{ind}=-L\dfrac{\Delta I}{\Delta t}\).

📌 So bulb \(B_2\) is delayed slightly, while the current reaches \(B_1\) (the resistor) immediately and it lights first.

\(L=\dfrac{\mu_0N^{2}A}{\ell}\): the piece has \(N^{\prime}=0.4N\), length \(\ell^{\prime}=0.4\ell\) and the same area.

\(\dfrac{L^{\prime}}{L}=\dfrac{(0.4N)^{2}/(0.4\ell)}{N^{2}/\ell}=0.4\Rightarrow L^{\prime}=0.4\times0.4=0.16\,H\).

📌 The field exists only inside the long solenoid, so the area crossed by the flux is that of the inner solenoid \(0.1\,m^{2}\) (not \(0.15\)).

\(M=\dfrac{\mu_0N_1N_2A}{\ell_1}=\dfrac{4\pi\times10^{-7}\times60\times6\times0.1}{0.2}=2.26\times10^{-4}\,H\).

📌 \(\Delta V_{ind}=-N\dfrac{\Delta\phi}{\Delta t}\): the emf equals minus the slope of the flux graph.

📌 Stage 1 (0–1 s): the flux increases so the emf is negative. Stage 2 (1–2 s): the flux is constant so the emf is zero. Stage 3 (2–5 s): the flux increases with a smaller slope so the emf is a smaller negative value. Stage 4 (5–6 s): the flux drops quickly so the emf is large and positive — this is graph D.

📌 \(\Delta V_{ind}=0\) when the flux is constant (zero slope).

📌 The stage \(B\Rightarrow C\) is horizontal (the flux is constant at \(60\times10^{-2}\)) whereas the slopes of the other stages are not zero.

📌 Before the magnet crosses the loop it approaches with its north pole upward, so the upward flux increases and a downward field is induced, i.e. a clockwise current.

📌 After crossing, the magnet moves away so the upward flux decreases and an upward field is induced, i.e. a counterclockwise current.

\(A=(0.10)^{2}=0.01\,m^{2}\).

\(|\Delta V_{ind}|=NA\left|\dfrac{\Delta B}{\Delta t}\right|=50\times0.01\times6=3\,V\).

📌 In figure A the flux through the ring changes as it moves, so eddy currents flow and oppose the motion (Lenz's law), stopping the pendulum.

📌 In figure B the ring moves parallel to the field so the flux through it does not change and no induced current flows; the pendulum continues without being slowed (neglecting air).

📌 An emf is produced when the wire cuts field lines: its velocity must be perpendicular to the field \(\vec{B}\) (from N to S, i.e. to the right) and to the wire at the same time.

📌 Motions C and D are parallel to the field and B is along the wire so no lines are cut; motion A (upward) is perpendicular to both so an emf is produced.

📌 A point at distance \(x\) from the axis has speed \(v=\omega x\) so \(d(\Delta V)=B\omega x\,dx\).

\(\Delta V_{ind}=\int_0^{L}B\omega x\,dx=B\omega\dfrac{L^{2}}{2}\) (or \(BL\bar{v}\) with the average speed \(\omega L/2\)).

📌 \(|\Delta V_{ind}|=BLv=0.6\times0.05\times0.01=3\times10^{-4}\,V\) while the loop enters (0 to 5 s) and while it leaves (15 to 20 s).

📌 It is zero in between (5–15 s) because the whole loop is inside the field and the flux is constant.

📌 While entering the inward flux increases so the emf is negative, while leaving it decreases so the emf is positive: the first pulse is negative and the second positive — graph D.

📌 \(\Delta V_{ind}=-L\dfrac{\Delta I}{\Delta t}\): the slope of the current graph is constant in each stage, so the emf is constant in each stage.

📌 While the current increases the emf is negative, and while it decreases it is positive with the same magnitude: a square wave, negative then positive — graph D.

📌 From the graph: at \(C\) (\(t=15\,s\)) the flux is \(60\times10^{-2}\) and at \(D\) (\(t=30\,s\)) it is zero.

\(\Delta V_{ind}=-N\dfrac{\Delta\phi}{\Delta t}=-30\times\dfrac{0-0.6}{15}=+1.2\,V\).

📌 A clockwise current (seen from the left) produces a field pointing to the right (away from the observer).

📌 The induced field opposes an increase of the leftward flux; the pole nearest the loop is the south pole (the flux at it points toward the pole, i.e. to the left), so the magnet must be brought toward the loop, i.e. to the right.

📌 The current in the wire is upward so its field at the loop (to its right) points into the page.

📌 A clockwise induced current produces an inward field, so it opposes a decrease of the inward flux, i.e. the loop moves away from the wire: in the \(X^{+}\) direction.

📌 Coil A: the magnet approaches it with its north pole so the flux increases, an opposing field is created and the current flows from \(D\) to \(C\) (by the grip rule).

📌 Coil B: the magnet moves away so the flux decreases, a field similar to the cause is created, i.e. opposite to that in A; the coils are alike so the current flows from \(X\) to \(Y\).

📌 The flux changes as the coil rotates so a sinusoidal emf is produced in the coil.

📌 The split rings (commutator) swap the connection of the coil to the resistor every half turn, so the current in the resistor keeps one direction while its value varies: a pulsating, always-positive current — graph A.

📌 With two full rings the connection is not reversed, so the current reverses its direction every half turn and its value varies sinusoidally.

📌 This is an alternating current — graph C.

📌 A current flows in the external circuit only when the sides of the coil cut the field lines, i.e. when their motion is not parallel to the field.

📌 In positions D and B the wires move parallel to the field so no current flows; in position C the current flows in the direction drawn (right-hand rule) and in position A it flows opposite to the direction drawn.

\(B=\mu_0\dfrac{N}{\ell}I\): \(I\) doubles and \(N/\ell\) doubles (since \(\ell\) is halved).

\(B^{\prime}=2\times2\times0.05=0.2\,T\).

📌 The surface is perpendicular to the field, so the field is parallel to the normal: \(\phi=BA=0.4\times0.05=0.02\,T\cdot m^{2}\).

📌 \(\phi=BA\cos\theta\) with \(\theta\) between the field and the normal; it vanishes when \(\theta=90^{\circ}\), i.e. when the field is parallel to the surface (no field lines pass through it).

\(|\Delta V|=A\dfrac{\Delta B}{\Delta t}=0.2\times\dfrac{0.5-0.1}{0.2}=0.4\,V\).

\(|\Delta V|=NA\dfrac{\Delta B}{\Delta t}=100\times0.01\times\dfrac{0.2}{0.1}=2\,V\).

📌 The magnetic field of the induced current opposes the change in flux (not the flux itself); this expresses energy conservation and explains the minus sign in Faraday's law.

📌 The flux toward the loop (away from the north pole) increases, so an opposing field is induced: the loop behaves as a north pole facing the magnet (repelling it).

📌 The north face of a loop is the one from which the current is seen counterclockwise.

\(\Delta V=BLv=0.3\times0.5\times4=0.6\,V\).

\(I=\dfrac{\Delta V}{R}=\dfrac{0.6}{3}=0.2\,A\).

📌 The speed is constant so the applied force equals the magnetic force on the current-carrying wire: \(F=BIL=0.3\times0.2\times0.5=0.03\,N\).

📌 Check: the mechanical power \(Fv=0.12\,W\) equals the electrical power \(I^{2}R=0.04\times3=0.12\,W\).

\(|\Delta V|=L\dfrac{\Delta I}{\Delta t}=0.5\times\dfrac{4}{0.2}=10\,V\).

\(\Delta V_{max}=NBA\omega=200\times0.5\times0.02\times100=200\,V\).

\(\Delta V_{max}=NBA\omega\) is directly proportional to \(\omega\), and the frequency \(f=\dfrac{\omega}{2\pi}\) is also directly proportional to it.

\(\Delta V=\dfrac{1}{2}B\omega L^{2}=\dfrac{1}{2}\times0.2\times40\times0.25=1.0\,V\).

\(\dfrac{d\phi}{dt}=0.04t=0.08\,V\) at \(t=2\,s\).

\(|\Delta V|=N\dfrac{d\phi}{dt}=50\times0.08=4\,V\).

\(|\Delta V|=N\dfrac{\Delta\phi}{\Delta t}=100\times\dfrac{0.04}{0.4}=10\,V\).

\(I=\dfrac{10}{10}=1\,A\).

📌 A current from the battery flows in the coil and produces a magnetic field around it.

📌 When the magnet approaches, the magnetic flux through the coil increases and a field opposite to the cause is induced (Lenz's law).

📌 This induced field is opposite to the field of the battery current, so an induced emf opposes the battery's emf and the current decreases.

📌 When the switch in the first circuit is opened its current decreases so the flux through the second coil decreases.

📌 In the second coil a field is induced that compensates the decrease, i.e. in the same direction as the field of the first coil (Lenz's law).

📌 This field is opposite to the field of the current flowing in the second coil itself, so the current of the second circuit decreases for a moment and then returns to its value.

📌 (a) Position C (leaving): the loop leaves the field so the inward flux decreases and a similar (inward) field is induced, i.e. a clockwise current.

📌 (b) Position B (completely inside the field): the flux is constant so no emf and no current are induced.

📌 In position A (entering) the inward flux increases so an opposing (outward) field is induced, i.e. a counterclockwise current.

📌 Figure 1: an electric generator, rotating counterclockwise, based on electromagnetic induction; the split rings reverse the connection every half turn so the current in the external circuit keeps one direction.

📌 Figure 2: an electric motor, rotating clockwise, based on the magnetic force on a current-carrying wire; the split rings reverse the current in the coil every half turn so the rotation continues in the same direction.

📌 (a) The wire is the source of the emf and the current leaves the positive pole in the external circuit; the drawn current leaves at B, so A is the negative pole and B the positive pole.

📌 (b) Inside the source the current flows from the negative to the positive pole: from A to B.

📌 (c) With the three-finger rule (thumb: motion of the wire downward, forefinger: the field, middle finger: current from A to B): the field is perpendicular into the page.

📌 (d) \(V_{ind}=BLv\sin\theta\) and the smallest field occurs when \(\sin\theta=1\): \(B=\dfrac{V_{ind}}{Lv}=\dfrac{0.15}{0.2\times3}=0.25\,T\).

📌 (a) The current rises from zero to \(10\,A\): \(\Delta V_{ind}=-L\dfrac{\Delta I}{\Delta t}\Rightarrow-2=-L\dfrac{10-0}{0.1}\) so \(L=0.02\,H\).

📌 (b) \(\Delta V_{ind}=-N\dfrac{\Delta\phi}{\Delta t}\Rightarrow-2=-8\dfrac{\phi-0}{0.1}\) so \(\phi=0.025\,T\cdot m^{2}\) (check: \(L=\dfrac{N\phi}{I}=\dfrac{8\times0.025}{10}=0.02\,H\)).

📌 (c) The statement is correct: \(L=-\Delta V_{ind}\Big/\dfrac{\Delta I}{\Delta t}\). When the current rises \(\Delta I\gt 0\) and \(\Delta V_{ind}\lt 0\) so \(L=-\dfrac{(-)}{(+)}=+\); when it falls \(\Delta I\lt 0\) and \(\Delta V_{ind}\gt 0\) so \(L=-\dfrac{(+)}{(-)}=+\). The sign of the emf follows the sign of \(\Delta I\) and \(L\) is always positive.