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BaLLatris [955]
3 years ago
7

If a wire lies withina magnetic field what must be true for the magnetic field to produce an electric current in the wire

Physics
1 answer:
BigorU [14]3 years ago
5 0

Answer:

The magnetic field through the wire must be changing

Explanation:

According to Faraday's law, the induced emf, ε in a metallic conductor is directly proportional to the rate of change of magnetic flux,Φ  through it. This is stated mathematically as ε = dΦ/dt.

Now for the wire, the magnetic flux through it is given by Φ = ABcosθ where A = cross-sectional area of wire, B = magnetic field and θ = angle between A and B.

So, dΦ/dt = dABcosθ/dt

Since A and B are constant,

dΦ/dt = ABdcosθ/dt = -(dθ/dt)ABsinθ

Since dθ/dt implies a change in the angle between A and B, since A is constant, it implies that B must be rotating.

So, <u>for an electric current (or voltage) to be produced in the wire, the magnetic field must be rotating or changing</u>.

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- thumb: force --> due north
However, we have to take into account that the electron has negative charge, therefore we have to take the opposite direction: so, the magnetic force is directed southwards, and the electrons are deflected due south.

b) From the kinetic energy of the electrons, we can find their velocity by using
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v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 2.20 \cdot 10^{-15} J}{9.1 \cdot 10^{-31} kg} }=6.95 \cdot 10^7 m/s

The Lorentz force due to the magnetic field provides the centripetal force that deflects the electrons:
qvB = m \frac{v^2}{r}
where
q is the electron charge
v is the speed
B is the magnetic field strength
m is the electron mass
r is the radius of the trajectory
By re-arranging the equation, we find the radius r:
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