The work done to transport an electron from the positive to the negative terminal is 1.92×10⁻¹⁹ J.
Given:
Potential difference, V = 1.2 V
Charge on an electron, e = 1.6 × 10⁻¹⁹ C
Calculation:
We know that the work done to transport an electron from the positive to the negative terminal is given as:
W.D = (Charge on electron)×(Potential difference)
= e × V
= (1.6 × 10⁻¹⁹ C)×(1.2 V)
= 1.92 × 10⁻¹⁹ J
Therefore, the work done in bringing the charge from the positive terminal to the negative terminal is 1.92 × 10⁻¹⁹ J.
Learn more about work done on a charge here:
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Answer:
13.78 mT
Explanation:
The peak voltage ε = ωNAB where ω = angular speed of coil = 1500 rpm = 1500 × 2π/60 rad/s = 50π rad/s = 157.08 rad/s, N = number of turns of coil = 250, A = area of coil = πr² where r = radius of coil = 10 cm = 0.10 m,
A = π(0.1 m)² = 0.03142 m² and B = magnetic field strength
So,
B = ε/ωNA
substituting the values of the variables into the equation given that ε = 17 V
So, B = ε/ωNA
B = 17 V/(157.08 rad/s × 250 turns × 0.03142 m²)
B = 17 V/(1233.8634 rad-turns-m²/s)
B = 0.01378 T
B = 13.78 mT
I believe it is False, only because the plane is Frictionless. Hope this helps, good luck.
Use the impulse-momentum theorem.

Substitute your known values:

Hope this helps!