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exis [7]
2 years ago
8

The magnetic field strength within a long solenoid isb=4.0t t, where t is time in seconds. if the radius of thesolenoid is1.0 cm

, what is the magnitude of the non-coulomb force that acts on a6.0 c charge a distance6.0 cmfrom the axis of the solenoid?
Physics
1 answer:
densk [106]2 years ago
3 0
The current problem can solved by using Faraday's law of induction which is actually one of the Maxwell's equations. It is stated as follows:
F=qE

Where
F = Non-coulomb force
q = 6.0 C
E = (dB/dt)*r^2*(2/R); In which dB/dt = 4 T; r = 1 cm = 0.01 m, R = 6.0 cm = 0.06 m, E = Newtons/Coulob
Substituting;

E = 4*0.01^2*(2/0.06) = 1/75

Therefore,
F = qE = 6*1/75 = 0.08 N
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Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote
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Answer:

If all these three charges are positive with a magnitude of 1.8 \times 10^{-8}\; \rm C each, the electric potential at the midpoint of segment \rm AB would be approximately 8.3 \times 10^{3}\; \rm V.

Explanation:

Convert the unit of the length of each side of this triangle to meters: 10\; \rm cm = 0.10\; \rm m.

Distance between the midpoint of \rm AB and each of the three charges:

  • d({\rm A}) = 0.050\; \rm m.
  • d({\rm B}) = 0.050\; \rm m.
  • d({\rm C}) = \sqrt{3} \times (0.050\; \rm m).

Let k denote Coulomb's constant (k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}.)

Electric potential due to the charge at \rm A: \displaystyle \frac{k\, q}{d({\rm A})}.

Electric potential due to the charge at \rm B: \displaystyle \frac{k\, q}{d({\rm B})}.

Electric potential due to the charge at \rm A: \displaystyle \frac{k\, q}{d({\rm C})}.

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of \rm AB due to all these three charges  would be:

\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}.

4 0
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Answer:

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Explanation:

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A 4kg block sitting on the floor, how much potential energy does it have?
prohojiy [21]

Well, there you have a very important principle wrapped up in that question.

There's actually no such thing as a real, actual amount of potential energy.
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Here's the formula for potential energy:    PE = (mass) x (gravity) x (<em><u>height</u></em><u>)</u> .

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Let's say that gravity is 9.8 m/s² .

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