(a) What is a magnetic field?
(b) With the aid of a labelled diagram describe an experiment to show that a magnetic field exists around a straight wire carrying current.
(c) A 40 \(\mu\)F capacitor in series with a 40 Q resistor is connected to a 100 V, 50 Hz a.c. supply.
(i) Draw a circuit diagram of the arrangement
(ii) Calculate the: I. impedance in the circuit; II. current in the circuit III. potential difference across the capacitor.
(b) With the aid of a labelled diagram describe an experiment to show that a magnetic field exists around a straight wire carrying current.
(c) A 40 \(\mu\)F capacitor in series with a 40 Q resistor is connected to a 100 V, 50 Hz a.c. supply.
(i) Draw a circuit diagram of the arrangement
(ii) Calculate the: I. impedance in the circuit; II. current in the circuit III. potential difference across the capacitor.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option n
Explanation:
(a) A magnetic field is a region or space where a magnetic force is experienced.
(b)
Connect a straight wire to the circuit as shown above, sprinkle iron fillings on the cardboard paper round the straight wire; and tap gently. Then close the key and gently tap again, the iron fillings form a concentric circles round the wire. This shows that there is magnetic field round a current carrying conductor.
(c)(i)
\(x_c = \frac{1}{2 \pi fc} = \frac{1}{2 \pi \times 50(40 \times 10^{-6})}\)
= 79.567\(\Omega\)
z = \(\sqrt{R^2 + X^2_c} = \sqrt{40^2 + 79,580^2}\)
(ii) I = \(\frac{v}{2} = \frac{100}{89.1} = 1.12A\)
(iii) V = IX\(_c\) = 1.12 x 79.567
= 89.1V
(a) A magnetic field is a region or space where a magnetic force is experienced.
(b)
Connect a straight wire to the circuit as shown above, sprinkle iron fillings on the cardboard paper round the straight wire; and tap gently. Then close the key and gently tap again, the iron fillings form a concentric circles round the wire. This shows that there is magnetic field round a current carrying conductor.
(c)(i)
\(x_c = \frac{1}{2 \pi fc} = \frac{1}{2 \pi \times 50(40 \times 10^{-6})}\)
= 79.567\(\Omega\)
z = \(\sqrt{R^2 + X^2_c} = \sqrt{40^2 + 79,580^2}\)
(ii) I = \(\frac{v}{2} = \frac{100}{89.1} = 1.12A\)
(iii) V = IX\(_c\) = 1.12 x 79.567
= 89.1V