When a positively charged conductor is placed near a candle flame, the flame spreads out as shown in the diagram above. Explain this observation.
(b) A proton moving with a speed of 5.0 x 10\(^{5}\) ms\(^{-1}\) enters a magnetic field of flux density 0.2 T at an angle of 30° to the field. Calculate the magnitude of the magnetic fcrce exerted on the proton. [Proton charge = 1.6 x 10\(^{-19}\) C]
The diagram above illustrates a 9.0 V battery of internal resistance 0.5 \(\Omega\) connected to two resistors of values 2.0 \(\Omega\) and R \(\Omega\). A\(_1\) A\(_2\) and A\(_3\) are ammeters of negligible internal resistances. If Al reads 4.0 A, calculate the:
(i) equivalent resistance of the combined resistors 2.0 \(\Omega\) and R \(\Omega\);
(ii) currents through A\(_1\) and A\(_3\) ; (iii) value of R.
(a) Why the candle flame spreads out near the positive conductor
A candle flame is a region of hot, ionised gas: the burning gases contain positive ions, negative ions and free electrons. The pointed positively charged conductor sets up a strong electric field around itself, which exerts forces on these charges.
- Negative ions and electrons in the flame are attracted towards the positive conductor.
- Positive ions are repelled away from it.
The repelled positive ions collide with, and drag along, the surrounding neutral air molecules, producing a stream of moving air called an electric wind. This wind blows the flame gases outward, so the flame is pushed away and appears to spread out, exactly as shown in the diagram.
(b) Magnetic force on the proton
The magnetic force on a charge moving at an angle \(\theta\) to a field is
\[ F = qvB\sin\theta \]
With \(q = 1.6\times10^{-19}\,\text{C}\), \(v = 5.0\times10^{5}\,\text{ms}^{-1}\), \(B = 0.2\,\text{T}\) and \(\theta = 30^{\circ}\):
\[ F = (1.6\times10^{-19})(5.0\times10^{5})(0.2)(\sin 30^{\circ}) \]
\[ F = (1.6\times10^{-19})(5.0\times10^{5})(0.2)(0.5) = 8.0\times10^{-15}\,\text{N} \]
(c) The battery-and-resistors circuit
EMF \(E = 9.0\,\text{V}\), internal resistance \(r = 0.5\,\Omega\), and the main-line ammeter \(A_1\) reads the total current \(I = 4.0\,\text{A}\). The 2.0 \(\Omega\) and \(R\) resistors are joined in parallel.
(i) Equivalent resistance of the combined resistors
Total resistance of the circuit is
\[ R_{total} = \frac{E}{I} = \frac{9.0}{4.0} = 2.25\,\Omega \]
This total is the internal resistance in series with the parallel combination, so
\[ R_{eq} = R_{total} - r = 2.25 - 0.5 = 1.75\,\Omega \]
(ii) Currents through \(A_1\) and \(A_3\)
\(A_1\) is in the main line, so it reads the total current, \(I_1 = 4.0\,\text{A}\).
The p.d. across the parallel combination is
\[ V = I\,R_{eq} = 4.0 \times 1.75 = 7.0\,\text{V} \]
The current in the 2.0 \(\Omega\) branch (read by \(A_2\)) is
\[ I_2 = \frac{V}{2.0} = \frac{7.0}{2.0} = 3.5\,\text{A} \]
By Kirchhoff's current rule, the current through the \(R\) branch (ammeter \(A_3\)) is
\[ I_3 = I_1 - I_2 = 4.0 - 3.5 = 0.5\,\text{A} \]
(iii) Value of R
\[ R = \frac{V}{I_3} = \frac{7.0}{0.5} = 14\,\Omega \]
Check: \(\dfrac{1}{R_{eq}} = \dfrac{1}{2.0} + \dfrac{1}{14} = \dfrac{7+1}{14} = \dfrac{8}{14}\), giving \(R_{eq} = 1.75\,\Omega\), which agrees with part (i).
