(a)
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]
(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.
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]
(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.
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Correct Answer: Option n
Explanation:
(a) The positively charged conductor attracts the negative charges in the air and repels the positive charges i.e. the candle flame ionizes the air around it.
= 1.6 x 10\(^{-19}\) X 5.0 X 10\(^5\) x 0.2 x Sin 30°
= 0.8 x 10\(^{-1}\)
(c)(i)E = I (Rc + r)
9 = 4(Rc + 0.5)
R: = 9/4-0.5
= 1.75Ω
(ii) Lost volt = Ir = 4.0 x 0.5 =2.0 V
V oltage across 2Ω and RΩ
V = 9.0 - 2.0 = 7.0V
Current in A\(_2\) = I\(_2\) = \(\frac{2}{VR}\) = 7/2 = 3.5A
Current in A\(_3\) = I\(_3\) = V/R = 7/14 = 0.5A
OR
A\(_3\) = A\(_1\) - A\(_2\)
= 4.0 - 3.5 = 0.5A
(iii) 1/RC = \(\frac{1}{R}\) + \(\frac{1}{R_2}\)
1/1.75 = \(\frac{1}{R}\) + \(\frac{1}{2}\)\(\frac{1}{R}\) = 1/1.75 - \(\frac{1}{2}\)
R = 14Ω
(a) The positively charged conductor attracts the negative charges in the air and repels the positive charges i.e. the candle flame ionizes the air around it.
= 1.6 x 10\(^{-19}\) X 5.0 X 10\(^5\) x 0.2 x Sin 30°
= 0.8 x 10\(^{-1}\)
(c)(i)E = I (Rc + r)
9 = 4(Rc + 0.5)
R: = 9/4-0.5
= 1.75Ω
(ii) Lost volt = Ir = 4.0 x 0.5 =2.0 V
V oltage across 2Ω and RΩ
V = 9.0 - 2.0 = 7.0V
Current in A\(_2\) = I\(_2\) = \(\frac{2}{VR}\) = 7/2 = 3.5A
Current in A\(_3\) = I\(_3\) = V/R = 7/14 = 0.5A
OR
A\(_3\) = A\(_1\) - A\(_2\)
= 4.0 - 3.5 = 0.5A
(iii) 1/RC = \(\frac{1}{R}\) + \(\frac{1}{R_2}\)
1/1.75 = \(\frac{1}{R}\) + \(\frac{1}{2}\)\(\frac{1}{R}\) = 1/1.75 - \(\frac{1}{2}\)
R = 14Ω