Please help to solve 11-20 Step by step

Here are the solutions for questions 11-20.

Question 11: Step 1: Identify the given values and the formula for motional EMF. The speed of the train is $v = 10 \text{ m/s}$. The magnetic flux density is $B = 10 \text{ μT} = 10 \times 10^{-6} \text{ T}$. The length of the axle is $L = 1.0 \text{ m}$. The magnetic field is perpendicular to the velocity and the axle, so $\sin\theta = 1$. The formula for motional EMF is $\mathcal{E} = B L v \sin\theta$.

Step 2: Substitute the values into the formula and calculate the EMF. $\mathcal{E} = (10 \times 10^{-6} \text{ T}) \times (1.0 \text{ m}) \times (10 \text{ m/s}) \times 1$ $\mathcal{E} = 100 \times 10^{-6} \text{ V}$ $\mathcal{E} = 1.0 \times 10^{-4} \text{ V}$ The induced emf is $\boxed{\text{*}1.0 \times 10^{-4} \text{ V*}}$.

Question 12: Step 1: Identify the given values and the formula for induced EMF by Faraday's Law. Number of turns $N = 5$. Initial magnetic flux $\Phi_1 = 1.0 \text{ mWb} = 1.0 \times 10^{-3} \text{ Wb}$. Final magnetic flux $\Phi_2 = 2.0 \text{ mWb} = 2.0 \times 10^{-3} \text{ Wb}$. Time interval $\Delta t = 1.0 \text{ s}$. The magnitude of the induced EMF is given by $|\mathcal{E}| = N \frac{|\Delta\Phi|}{|\Delta t|}$.

Step 2: Calculate the change in magnetic flux. $\Delta\Phi = \Phi_2 - \Phi_1 = (2.0 \times 10^{-3} \text{ Wb}) - (1.0 \times 10^{-3} \text{ Wb}) = 1.0 \times 10^{-3} \text{ Wb}$

Step 3: Substitute the values into the EMF formula and calculate. $|\mathcal{E}| = 5 \times \frac{1.0 \times 10^{-3} \text{ Wb}}{1.0 \text{ s}}$ $|\mathcal{E}| = 5.0 \times 10^{-3} \text{ V}$ The magnitude of the induced emf is $\boxed{\text{*}5.0 \times 10^{-3} \text{ V*}}$.

Question 13: Step 1: Identify the given values and the formula for motional EMF. Wing span $L = 10 \text{ m}$. Speed of the aircraft $v = 100 \text{ m/s}$. Vertical component of Earth's magnetic field $B = 10 \text{ μT} = 10 \times 10^{-6} \text{ T}$. The wing span is perpendicular to both the velocity and the vertical magnetic field, so $\sin\theta = 1$. The formula for motional EMF is $\mathcal{E} = B L v \sin\theta$.

Step 2: Substitute the values into the formula and calculate the EMF. $\mathcal{E} = (10 \times 10^{-6} \text{ T}) \times (10 \text{ m}) \times (100 \text{ m/s}) \times 1$ $\mathcal{E} = 10000 \times 10^{-6} \text{ V}$ $\mathcal{E} = 1.0 \times 10^{-2} \text{ V}$ The induced emf is $\boxed{\text{*}1.0 \times 10^{-2} \text{ V*}}$.

Question 14: Step 1: Identify the given values and the formula for induced EMF by Faraday's Law. Number of turns $N = 100$. Initial magnetic flux $\Phi_1 = 2 \text{ mWb} = 2 \times 10^{-3} \text{ Wb}$. Final magnetic flux $\Phi_2 = 0 \text{ Wb}$. Time interval $\Delta t = 1.0 \text{ s}$. The magnitude of the average induced EMF is given by $|\mathcal{E}| = N \frac{|\Delta\Phi|}{|\Delta t|}$.

Step 2: Calculate the change in magnetic flux. $\Delta\Phi = \Phi_2 - \Phi_1 = 0 \text{ Wb} - (2 \times 10^{-3} \text{ Wb}) = -2 \times 10^{-3} \text{ Wb}$

Step 3: Substitute the values into the EMF formula and calculate. $|\mathcal{E}| = 100 \times \frac{|-2 \times 10^{-3} \text{ Wb}|}{1.0 \text{ s}}$ $|\mathcal{E}| = 100 \times (2 \times 10^{-3} \text{ V})$ $|\mathcal{E}| = 0.2 \text{ V}$ The average induced emf is $\boxed{\text{*}0.2 \text{ V*}}$.

Question 15: Step 1: Compare the given equation with the general form of alternating EMF. The given equation for emf is $\mathcal{E} = 20 \sin (100 \pi t)$. The general form for alternating emf is $\mathcal{E} = \mathcal{E}_0 \sin(\omega t)$, where $\mathcal{E}_0$ is the peak emf and $\omega$ is the angular frequency.

Step 2: Identify the peak value from the equation. By comparing the two equations, the peak value $\mathcal{E}_0$ is the coefficient of the sine function. $\mathcal{E}_0 = 20 \text{ V}$ The peak value $\mathcal{E}_0$ is $\boxed{\text{*}20 \text{ V*}}$.

Question 16: Step 1: Identify the given RMS voltage and the relationship between RMS and peak voltage. The RMS voltage of the main supply is $V_{rms} = 230 \text{ V}$. The relationship between RMS voltage ($V_{rms}$) and peak voltage ($V_{peak}$) for an AC supply is $V_{rms} = \frac{V_{peak}}{\sqrt{2}}$.

Step 2: Rearrange the formula to solve for peak voltage and substitute the value. $V_{peak} = V_{rms} \times \sqrt{2}$ $V_{peak} = 230 \text{ V} \times \sqrt{2}$ $V_{peak} \approx 230 \text{ V} \times 1.414$ $V_{peak} \approx 325.22 \text{ V}$ The peak voltage is $\boxed{\text{*}325 \text{ V*}}$.

Question 17: Step 1: Identify the given current equation and resistance, and the formulas for mean power. The alternating current is $I = 3 \sin (50t)$. From this, the peak current is $I_{peak} = 3 \text{ A}$. The resistive load is $R = 10 \text{ Ω}$. The mean power dissipated in a resistive load is $P_{mean} = I_{rms}^2 R$.

Step 2: Calculate the RMS current from the peak current. The relationship between RMS current ($I_{rms}$) and peak current ($I_{peak}$) is $I_{rms} = \frac{I_{peak}}{\sqrt{2}}$. $I_{rms} = \frac{3 \text{ A}}{\sqrt{2}}$

Step 3: Substitute the RMS current and resistance into the mean power formula. $P_{mean} = \left(\frac{3}{\sqrt{2}}\right)^2 \times 10 \text{ Ω}$ $P_{mean} = \frac{9}{2} \times 10 \text{ W}$ $P_{mean} = 4.5 \times 10 \text{ W}$ $P_{mean} = 45 \text{ W}$ The mean power dissipated is $\boxed{\text{*}45 \text{ W*}}$.

Question 18: Step 1: Identify the given power and RMS voltage, and the formulas for power and peak current. Power of the heater $P = 4.0 \text{ kW} = 4000 \text{ W}$. RMS voltage of the supply $V_{rms} = 240 \text{ V}$. The power in an AC circuit is $P = V_{rms} I_{rms}$. The relationship between RMS current and peak current is $I_{peak} = I_{rms} \sqrt{2}$.

Step 2: Calculate the RMS current. $I_{rms} = \frac{P}{V_{rms}}$ $I_{rms} = \frac{4000 \text{ W}}{240 \text{ V}}$ $I_{rms} = \frac{400}{24} \text{ A} = \frac{50}{3} \text{ A} \approx 16.67 \text{ A}$

Step 3: Calculate the peak current. $I_{peak} = I_{rms} \sqrt{2}$ $I_{peak} = \frac{50}{3} \text{ A} \times \sqrt{2}$ $I_{peak} \approx 16.67 \text{ A} \times 1.414$ $I_{peak} \approx 23.57 \text{ A}$ The peak current drawn is $\boxed{\text{*}23.6 \text{ A*}}$.

Question 19: Step 1: Identify the given voltages and secondary turns, and the transformer turns ratio formula. Primary voltage $V_p = 220 \text{ V}$. Secondary voltage $V_s = 7.5 \text{ V}$. Number of turns on the secondary coil $N_s = 140$ turns. For an ideal transformer, the ratio of voltages is equal to the ratio of turns: $\frac{V_p}{V_s} = \frac{N_p}{N_s}$.

Step 2: Rearrange the formula to solve for the number of turns on the primary coil ($N_p$) and substitute the values. $N_p = N_s \times \frac{V_p}{V_s}$ $N_p = 140 \text{ turns} \times \frac{220 \text{ V}}{7.5 \text{ V}}$ $N_p = 140 \times 29.333...$ $N_p \approx 4106.67$ Since the number of turns must be an integer, we round to the nearest whole number. The number of turns on the primary coil is $\boxed{\text{*}4107 \text{ turns*}}$.

Question 20: Step 1: Identify the given voltages and secondary current, and the power conservation formula for an ideal transformer. Primary voltage $V_p = 240 \text{ V}$. Secondary voltage $V_s = 15 \text{ V}$. Secondary current $I_s = 2.0 \text{ A}$. For an ideal transformer, the power in the primary coil equals the power in the secondary coil: $P_p = P_s$. This can be written as $V_p I_p = V_s I_s$.

Step 2: Rearrange the formula to solve for the primary current ($I_p$) and substitute the values. $I_p = \frac{V_s I_s}{V_p}$ $I_p = \frac{(15 \text{ V}) \times (2.0 \text{ A})}{240 \text{ V}}$ $I_p = \frac{30 \text{ W}}{240 \text{ V}}$ $I_p = \frac{1}{8} \text{ A}$ $I_p = 0.125 \text{ A}$ The current drawn from the primary coil is $\boxed{\text{*}0.125 \text{ A*}}$.