Question

1. (a) in a tabular form classify the following substances as ferromagnetic or diamagnetic materials: iron, nickel, cobalt, lead, gadolinium, mercury. 3 marks (b) two capacitors 8 μf and 4 μf are connected in parallel and placed in series with a 6 μf capacitor. (i) draw the circuit - diagram for the arrangement. (ii) calculate the effective capacitance in the circuit. 5 marks (c) (i) a step - up transformer is designed to operate from a 20 v supply to deliver 200 v. if the efficiency of the transformer is 80 %. calculate the current in the primary coil when the output terminals are connected to a 300 v and 150 w lamp. (ii) a charged particle travelling at a speed of 7.0×10^6 m s^(-1) enters at right - angle into a magnetic field of strength 0.30 t. if it moves in a circular path of radius 25 cm in the field, calculate the ratio of the charge, q, to the mass, m, of the particle. 7 marks

Explanation:

Step1: Analyze capacitor - parallel combination

For two capacitors \(C_1 = 8\ pF\) and \(C_2=1\ pF\) in parallel, the equivalent capacitance of the parallel - combination \(C_{p}\) is given by \(C_{p}=C_1 + C_2\).

$$C_{p}=8\ pF+ 1\ pF=9\ pF$$

Step2: Analyze series combination

This parallel - combination \(C_{p}\) is in series with \(C_3 = 6\ pF\). The formula for the equivalent capacitance \(C_{eq}\) of two capacitors \(C_{p}\) and \(C_3\) in series is \(\frac{1}{C_{eq}}=\frac{1}{C_{p}}+\frac{1}{C_{3}}\).
Substitute \(C_{p}=9\ pF\) and \(C_3 = 6\ pF\) into the formula: \(\frac{1}{C_{eq}}=\frac{1}{9}+\frac{1}{6}=\frac{2 + 3}{18}=\frac{5}{18}\).
Then \(C_{eq}=\frac{18}{5}=3.6\ pF\).

For the transformer part:
The efficiency formula of a transformer is \(\eta=\frac{P_{out}}{P_{in}}\), where \(P_{out}=V_{s}I_{s}\) and \(P_{in}=V_{p}I_{p}\). Given \(V_{p} = 20\ V\), \(V_{s}=200\ V\) and \(\eta = 0.8\). Let the power in the secondary be \(P_{s}\). If we assume a load resistance \(R\) in the secondary, \(P_{s}=\frac{V_{s}^{2}}{R}\). Also, \(\eta=\frac{V_{s}I_{s}}{V_{p}I_{p}}\).
We know that for an ideal - like transformer (with efficiency considered), \(\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}\) and \(\eta=\frac{V_{s}I_{s}}{V_{p}I_{p}}\). Since \(P_{s}=V_{s}I_{s}\) and \(P_{p}=V_{p}I_{p}\), and assuming the load power \(P_{s}\) is known (not given here, but we can also use the turns - ratio and efficiency relationship). If we assume the power in the secondary is \(P_{s}\), then \(P_{in}=\frac{P_{s}}{\eta}\). And \(I_{p}=\frac{P_{in}}{V_{p}}=\frac{P_{s}}{\eta V_{p}}\). If we assume the secondary is connected to a load such that \(P_{s}\) is the power dissipated in the load. But if we assume a simple case of power transfer, and since \(\eta=\frac{V_{s}I_{s}}{V_{p}I_{p}}\), we can also use the fact that for a step - up transformer \(\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}\). Let's assume the power in the secondary is \(P_{s}\). We know that \(P_{s}=V_{s}I_{s}\) and \(P_{p}=V_{p}I_{p}\), so \(I_{p}=\frac{V_{s}I_{s}}{\eta V_{p}}\). If we assume the secondary is open - circuit (no load for simplicity of finding the current relationship based on voltage and efficiency), we know that \(\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}\) and \(\eta=\frac{V_{s}I_{s}}{V_{p}I_{p}}\). Given \(V_{p} = 20\ V\), \(V_{s}=200\ V\) and \(\eta = 0.8\), we have \(I_{p}=\frac{V_{s}I_{s}}{\eta V_{p}}\). If we assume \(I_{s}\) is the current in the secondary. For a step - up transformer, \(\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}\). The power in the secondary \(P_{s}=V_{s}I_{s}\) and power in the primary \(P_{p}=V_{p}I_{p}\), so \(I_{p}=\frac{V_{s}I_{s}}{\eta V_{p}}\). If we assume the secondary is connected to a \(150\ W\) lamp at \(300\ V\), then \(I_{s}=\frac{P_{s}}{V_{s}}=\frac{150}{300}=0.5\ A\). Then \(I_{p}=\frac{V_{s}I_{s}}{\eta V_{p}}=\frac{200\times0.5}{0.8\times20}=\frac{100}{16}=6.25\ A\).

For the charged - particle in a magnetic field part:
The force on a charged particle moving in a magnetic field \(F = qvB\) provides the centripetal force \(F_{c}=\frac{mv^{2}}{r}\). So \(qvB=\frac{mv^{2}}{r}\), and we can solve for \(\frac{q}{m}\).

$$ \frac{q}{m}=\frac{v}{Br}$$

Given \(v = 7.0\times10^{6}\ m/s\), \(B = 0.30\ T\) and \(r=25\ cm = 0.25\ m\).

$$ \frac{q}{m}=\frac{7.0\times 10^{6}}{0.30\times0.25}=\frac{7.0\times10^{6}}{0.075}\approx9.33\times 10^{7}\ C/kg$$

Answer:

For the capacitor part, the effective capacitance is \(3.6\ pF\). For the transformer part, if the secondary is connected to a \(150\ W\) lamp at \(300\ V\), the current in the primary coil is \(6.25\ A\). For the charged - particle part, \(\frac{q}{m}\approx9.33\times 10^{7}\ C/kg\)