There are two types of generators, namely direct current (DC) or dynamo generators and alternating current (AC) generators or alternators.
The generator works based on the principle of electromagnetic induction, namely by rotating the coil in a magnetic field so that an induced emf is generated.
Meanwhile, a transformer or transformer is a device that is able to change the AC voltage by transferring electrical energy by induction through the primary coil to the secondary coil.
The transformer produces an emf in the secondary coil due to the changing magnetic field due to the flow of alternating current in the primary coil. The current is then induced by the malleable iron into the secondary coil.
Transformers are divided into two types, namely step-up and step-down transformers.
The step-up transformer functions to increase the source AC voltage, while the step-down transformer functions to decrease the source AC voltage.
Physics Problems And Solution - Electromagnetic induction is a symptom of the induced electromotive force on a conductor because there is a change in the magnetic flux that it surrounds. The electric current that arises due to the induced emf is called the induced current.
Electromagnetic Induction Problems and Solutions
Problems 1. Among the following factors:
(1) number of turns of coil
(2) rate of change of magnetic flux
(3) magnetic field direction
What affects the induced emf in the coil is...
Physics Solutions:
Faraday's law formula: \begin{align*}\varepsilon &=-N\frac{d\Phi }{dt}\end{align*} The induced emf is affected by the rate of change of the magnetic flux and the number of turns.
Problems 2. A generator produces an induced emf voltage of . If the generator coil is changed so that the number of turns is doubled from the original, and the flux rate is three times the original, the ratio of the current emf to the original emf is.....
Physics Solutions:\begin{align*}\frac{\varepsilon _1}{\varepsilon _2}&=\frac{N_1\frac{d\Phi _1}{dt}}{N_2\frac{d\Phi _2 }{dt}} \\\frac{\varepsilon _1}{\varepsilon _2}&=\frac{N_1.\Phi _1}{N_2.\Phi _2}\\\frac{\varepsilon _1}{\varepsilon _2 }&=\frac{N_1.\Phi _1}{2N_1.3\Phi _1}=\frac{1}{6}\end{align*}
Problems 3. A conductor 20 cm long is moved across a 5T magnetic field at a speed of 10 m/s. The magnitude of the induced emf between the two ends of the conductor is...
Physics Solutions:\begin{align*}\varepsilon &=Blv\sin \theta \\\varepsilon &=5(0,2)10\sin 90^o \\\varepsilon &=10V\end{align*}
Problems 4. If the magnetic induction B = 4 T and the wire PQ is shifted to the right with a speed of V, it turns out that the electric power at R is 25 watts.
Determine the magnitude of the speed V if the length of the wire PQ = 25 cm and the resistance R = 4 ohms.
Physics Solutions:\begin{align*}P&=\frac{v^{2}}{R}=\frac{(Blv)^{2}}{R}\\25&=\frac{(4( 0.25)v)^{2}}{4}\\v&=\sqrt{100}\\v&=10ms^{-2}\end{align*}
Problems 5. A coil with a cross-sectional area of 100 cm2, resistance 4 ohm with a number of turns 400 is in a magnetic field whose direction is parallel to the axis of the coil. The magnitude of the magnetic induction varies according to the equation B = 10-4 sin 2000t in SI units. Then the maximum induced current in the coil is...
Physics Solutions:\begin{align*}i&=\frac{\varepsilon }{R}\\i&=\frac{NA\frac{dB}{dt} }{R} \\i&=\frac{400 (100.10^{_4})\frac{d(10^{-4} \sin 2000t)}{dt} }{4} \\i&=\frac{4.10^{-4}.2000 \cos 2000t}{ 4}\\i&=0,2A=200mA\end{align*}
Problems 6. Lamps L1 and L2 have the same data 1 V, 15 mW arranged in a circuit as shown.
Physics Solutions:\begin{align*}P&=\frac{v^{2}}{R}=\frac{(Blv)^{2}}{R}\\25&=\frac{(4( 0.25)v)^{2}}{4}\\v&=\sqrt{100}\\v&=10ms^{-2}\end{align*}
Problems 5. A coil with a cross-sectional area of 100 cm2, resistance 4 ohm with a number of turns 400 is in a magnetic field whose direction is parallel to the axis of the coil. The magnitude of the magnetic induction varies according to the equation B = 10-4 sin 2000t in SI units. Then the maximum induced current in the coil is...
Physics Solutions:\begin{align*}i&=\frac{\varepsilon }{R}\\i&=\frac{NA\frac{dB}{dt} }{R} \\i&=\frac{400 (100.10^{_4})\frac{d(10^{-4} \sin 2000t)}{dt} }{4} \\i&=\frac{4.10^{-4}.2000 \cos 2000t}{ 4}\\i&=0,2A=200mA\end{align*}
Problems 6. Lamps L1 and L2 have the same data 1 V, 15 mW arranged in a circuit as shown.
When the PQ wire is shifted to the right with a speed of 10 m/s, both lights are lit normally. Determine the magnitude of the Lorentz force acting on the PQ wire!
Physics Solutions:
Both lights are on normally, meaning total power (2x15mW) = 20mW at 1 Volt voltage. \begin{align*}i&=\frac{P}{V}\\i&=\frac{30m}{1}=30mA\\\varepsilon &=Blv\\Bl&=\frac{\varepsilon }{v} =\frac{1}{10}\\F&=(Bl)i=(\frac{1}{10})(30m)\\F&=3mN=3.10^{-3}N\end{align*}
Problems 7. A conductor rod 80 cm long rotates with a frequency of 5 Hz in a 0.3 T magnetic field.
The magnitude of the induced emf that occurs in the conductor is approx..!
Physics Solutions:
If the conductor is rotated once, then the area enclosed by it is:\begin{align*}A&=\pi R^2=\pi l^2\end{align*}Number of circles N = 1
Time spent: \begin{align*}t=T\end{align*}so \begin{align*}\varepsilon &=NB\frac{\Delta A}{\Delta t}\\\varepsilon &=B \left (\frac{\pi l^2}{nT} \right )\\\varepsilon &=B\left (\frac{\pi l^2}{T} \right ) \\\varepsilon &= \ pi Bf l^2\end{align*}then\begin{align*}\varepsilon &=\pi fBl^2\\\varepsilon &=(3,14)(5) (0,3)(0,8 )^2 \\\varepsilon &\approx 3\, V\end{align*}
Problems 8. A current of 2A flows in a coil of 400 turns giving rise to a flux of 10-4 weber across the cross section of the coil. If the current reverses direction within 0.08 seconds, then the self-induced emf that occurs is...
Physics Solutions:\begin{align*}\varepsilon &=N\frac{\Delta \Phi }{\Delta t}\\\varepsilon &=400(\frac{10^{-4} }{0, 08})\\\varepsilon &=0,5V\end{align*}
Problems 9. A coil has 100 turns and the inductance is 0.4 Henry. If there is a change in the current in the coil from 10A to 2A in 0.1 second, the self-induced emf in the coil is...
Physics Solutions:\begin{align*}\varepsilon &=L\frac{\Delta i }{\Delta t}\\\varepsilon &=0.4(\frac{8 }{0,1})\\\varepsilon &=32V\end{align*}
Problems 10. A solenoid has 500 turns and is 10 cm long. solenoid radius of 5 cm. The inductance of the solenoid is approx.
Physics Solutions:\begin{align*}L&=\frac{\mu _{o}N^{2}A}{l}\\L&=\frac{4\pi .10^{-7}( 500)^{2}\times \pi (5.10^{-2})^{2}}{0.01}\\L&\approx 25.10^{-3}H=25mH\end{align*}
13. An inductor has an inductance of 50 mH. In the inductor flows 10A current. The energy stored in the inductor is...
Physics Solutions:\begin{align*}W&=\frac{1}{2}Li^{2}\\W&=\frac{1}{2}(50.10^{-3})10^{2 }\\W&=2.5J\end{align*}
Problems 11. A coil with an inductance of 3.0 mH carries a time-varying electric current according to the equation $I = 10 \sin 100 \pi t$ Ampere. The maximum self-induced emf in the coil is approx.…
Physics Solutions:\begin{align*}\varepsilon &=L\frac{di}{dt}\\\varepsilon &=3m\frac{d(10 \sin 100 \pi t)}{dt}\\ \varepsilon &=3m.10.100\pi \cos 100 \pi t\\\varepsilon _{mak} &=3\pi V\approx 10V\end{align*}
Problems 12. The ratio of the number of turns of wire in the primary and secondary coils of a transformer is 1: 4. The input voltage and current are 10V and 2A, respectively. If the average power converted into heat in the transformer is 4W and the output voltage is 40V, the output current is:
Physics Solutions:\begin{align*}P_{s}&=P_{p}-P_{h}\\V_{s}.i_{s}&=V_{ip}.i_{p}-P_ {k}\\40.i_{s}&=10.2-4\\40.i_{s}&=16\\i_s&=\frac{16}{40}\\i_s&=0.4A\end{align*}
Physics Solutions:
Both lights are on normally, meaning total power (2x15mW) = 20mW at 1 Volt voltage. \begin{align*}i&=\frac{P}{V}\\i&=\frac{30m}{1}=30mA\\\varepsilon &=Blv\\Bl&=\frac{\varepsilon }{v} =\frac{1}{10}\\F&=(Bl)i=(\frac{1}{10})(30m)\\F&=3mN=3.10^{-3}N\end{align*}
Problems 7. A conductor rod 80 cm long rotates with a frequency of 5 Hz in a 0.3 T magnetic field.
Physics Solutions:
If the conductor is rotated once, then the area enclosed by it is:\begin{align*}A&=\pi R^2=\pi l^2\end{align*}Number of circles N = 1
Time spent: \begin{align*}t=T\end{align*}so \begin{align*}\varepsilon &=NB\frac{\Delta A}{\Delta t}\\\varepsilon &=B \left (\frac{\pi l^2}{nT} \right )\\\varepsilon &=B\left (\frac{\pi l^2}{T} \right ) \\\varepsilon &= \ pi Bf l^2\end{align*}then\begin{align*}\varepsilon &=\pi fBl^2\\\varepsilon &=(3,14)(5) (0,3)(0,8 )^2 \\\varepsilon &\approx 3\, V\end{align*}
Problems 8. A current of 2A flows in a coil of 400 turns giving rise to a flux of 10-4 weber across the cross section of the coil. If the current reverses direction within 0.08 seconds, then the self-induced emf that occurs is...
Physics Solutions:\begin{align*}\varepsilon &=N\frac{\Delta \Phi }{\Delta t}\\\varepsilon &=400(\frac{10^{-4} }{0, 08})\\\varepsilon &=0,5V\end{align*}
Problems 9. A coil has 100 turns and the inductance is 0.4 Henry. If there is a change in the current in the coil from 10A to 2A in 0.1 second, the self-induced emf in the coil is...
Physics Solutions:\begin{align*}\varepsilon &=L\frac{\Delta i }{\Delta t}\\\varepsilon &=0.4(\frac{8 }{0,1})\\\varepsilon &=32V\end{align*}
Problems 10. A solenoid has 500 turns and is 10 cm long. solenoid radius of 5 cm. The inductance of the solenoid is approx.
Physics Solutions:\begin{align*}L&=\frac{\mu _{o}N^{2}A}{l}\\L&=\frac{4\pi .10^{-7}( 500)^{2}\times \pi (5.10^{-2})^{2}}{0.01}\\L&\approx 25.10^{-3}H=25mH\end{align*}
13. An inductor has an inductance of 50 mH. In the inductor flows 10A current. The energy stored in the inductor is...
Physics Solutions:\begin{align*}W&=\frac{1}{2}Li^{2}\\W&=\frac{1}{2}(50.10^{-3})10^{2 }\\W&=2.5J\end{align*}
Problems 11. A coil with an inductance of 3.0 mH carries a time-varying electric current according to the equation $I = 10 \sin 100 \pi t$ Ampere. The maximum self-induced emf in the coil is approx.…
Physics Solutions:\begin{align*}\varepsilon &=L\frac{di}{dt}\\\varepsilon &=3m\frac{d(10 \sin 100 \pi t)}{dt}\\ \varepsilon &=3m.10.100\pi \cos 100 \pi t\\\varepsilon _{mak} &=3\pi V\approx 10V\end{align*}
Problems 12. The ratio of the number of turns of wire in the primary and secondary coils of a transformer is 1: 4. The input voltage and current are 10V and 2A, respectively. If the average power converted into heat in the transformer is 4W and the output voltage is 40V, the output current is:
Physics Solutions:\begin{align*}P_{s}&=P_{p}-P_{h}\\V_{s}.i_{s}&=V_{ip}.i_{p}-P_ {k}\\40.i_{s}&=10.2-4\\40.i_{s}&=16\\i_s&=\frac{16}{40}\\i_s&=0.4A\end{align*}
