Earth has a magnetic dipole moment of 8.0 × 1022 J/T. (a) What current would have to be produced in a single turn of wire extend
ing around Earth at its geomagnetic equator if we wished to set up such a dipole? Could such an arrangement be used to cancel out Earth's magnetism (b) at points in space well above Earth's surface or (c) on Earth's surface? Round your answer (a) to 3 significant digits.
1 answer:
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Answer:
t = 23.9nS
Explanation:
given :
Area A= 10 cm by 2 cm => 2 x 10^-2m x 10 x 10^-2m
distance d= 1mm=> 0.001
resistor R= 975 ohm
Capacitance can be calculated through the following formula,
C = (ε0 x A )/d
C = (8.85 x 10^-12 x (2 x 10^-2 x 10 x 10^-2))/0.001
C = 17.7 x 10^-12 (pico 'p' = 10^-12)
C = 17.7pF
the voltage between two plates is related to time, There we use the following formula of the final voltage
Vc = Vx (1-e^-(t/CR))
75 = 100 x (1-e^-(t/CR))
75/100 = (1-e^-(t/CR))
.75 = (1-e^-(t/CR))
.75 -1 = -e^-(t/CR)
-0.25 = -e^-(t/CR) --->(cancelling out the negative sign)
e^-(t/CR) = 0.25
in order to remove the exponent, take logs on both sides
-t/CR = ln (0.25)
t/CR = -ln(0.25)
t = -CR x ln (0.25)
t = -(17.7 x 10^-12 x 975) x (-1.38629)
t = 23.9 x
t = 23.9ns
Thus, it took 23.9ns for the potential difference between the deflection plates to reach 75 volts