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OLga [1]
2 years ago
7

How would the self-inductance, l, of a coil change if you would increase its radius by a factor of two and increase its length b

y a factor of two?
Physics
1 answer:
pochemuha2 years ago
3 0

The self-inductance of a coil will change by 8 times its original value by increasing its radius value by 2 and increasing the length of the coil by 2.

Self-Inductance: -

The definition of self-inductance is the induction of a voltage in a wire that carries current when the current in the wire is changing. In the instance of self-inductance, the circuit itself induces a voltage through the magnetic field produced by a changing current.

We know that the self-inductance of the coil is denoted by: -

L= µ *π*(r)^2*(N)^2*l

Where

L= Self-Inductance of the coil

µ= Magnetic Permeability Constant

r= Radius of the coil

l= Length of the coil

N= Number of turns of the coil

Here Self-inductance of the coil is directly proportional to the length of the coil and the square of the radius of the coil.

So,

On increasing the radius of the coil by a factor of 2 and the length of the coil by 2 the self-inductance of the coil increases by 8 times its original value.

Learn more about Self-Inductance here: -

" brainly.com/question/15293029 "

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The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

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The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

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