Moving a magnet inside of a coil of wire will induce a voltage in the coil. How is the voltage in the coil increased?
1 answer:
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Answer:
Circle
Explanation:
When a charged particle is in motion in a region with magnetic field, the particle experiences a force whose magnitude is given by
where
q is the charge
v is the velocity of the particle
B is the strength of the magnetic field
is the angle between the directions of v and B
In this problem, the velocity of the particle is perpendicular to the magnetic field, so
and the formula reduces to
Also, the direction of this force is perpendicular to the direction of motion of the particle. This means that as the charge moves in the region of the magnetic field, the force acting on it acts as a centripetal force: therefore, the particle will start moving by unifom circular motion, with constant speed (because the magnetic force does no work on the particle, since it is perpendicular to the direction of motion).
So, the path of the particle will be a circle.
A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. The change in the electric potential energy of the proton-field system when the proton travels to x = 2.50m is -3.40 × 10⁻¹⁶ J (Option B)
<h3 /><h3>How is the change in electric potential energy of the proton-field system calculated?</h3>- Work done on the proton =Negative of the change in the electric potential energy of the proton field
- In the given case, W = -qΔV
- -W = qΔV
- = qEcosθ
- Therefore, work done on the proton = -e(8.50×
N/C)(2.5m)(1)
- = -3.40×
J
- Any change in the potential energy indicates the work done by the proton.
- Therefore the positive sign shows that the potential energy increases when the proton does the work.
- The negative sign shows that the potential energy decreases when the proton does the work.
To learn more about electric potential energy, refer
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