jengoy03
Answered

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The magnetic field perpendicular to a single wire loop of diameter 10.0 cm decreases from 0.50 T to zero. The wire is made of copper and has a diameter of 2.0 mm and length 1.0 cm. How much charge moves through the wire while the field is changing?

I know how to do the calculations, but can someone please explain what is the 10cm diameter and 2mm diameter? Why is there one wire and two diameters? I understand this problem mathematically but not conceptually.


Sagot :

leena

Hi there!

We can begin by using Lenz's Law:
[tex]\epsilon = -N\frac{d\Phi _B}{dt}[/tex]

N = Number of Loops

Ф = Magnetic Flux (Wb)
t = time (s)

Also, we can rewrite this as:
[tex]\epsilon = -NA\frac{dB}{dt}[/tex]

A = Area (m²)

Since the area is constant, we can take it out of the derivative.

This is a single wire loop, so N = 1.

Now, we can develop an expression for the induced emf.

We can begin by solving for the area:

[tex]A = \pi r^2 \\\\d = r/2 r = 0.05cm \\\\A = \pi (0.05^2) = 0.007854 m^2[/tex]

We can also express dB/dt as:
[tex]\frac{dB}{dt} = \frac{\Delta B}{t} = \frac{0-0.5}{t} = \frac{-0.5}{t}[/tex]

Now, we can create an equation.

[tex]\epsilon = -(1)(0.007854)\frac{-0.5}{t} = \frac{0.003927}{t}[/tex]

To solve the system, we must now develop an expression for current given an emf and resistance.

Begin by calculating the resistance of the copper wire:
[tex]R = \frac{\rho L}{A}[/tex]

ρ = Resistivity of copper (1.72 * 10⁻⁸ Ωm)
L = Length of wire (0.01 m)

A = cross section area (m²)

Solve:
[tex]R = \frac{(1.72*10^{-8})(0.01)}{\pi (0.001^2)} = 5.475 * 10^{-5} \Omega m[/tex]

Now, we can use the following relation (Ohm's Law):

[tex]\epsilon = iR\\\\\epsilon = \frac{Q}{t}R[/tex]

*Since current is equivalent to Q/t.

Plug in the value of R and set the two equations equal to each other.

[tex]\frac{Q}{t}(5.475 * 10^{-5}) = \frac{0.003927}{t}[/tex]

Cancel out 't'.

[tex]Q (5.475 * 10^{-5}) = 0.003927 \\\\Q = \frac{0.003927}{5.475*10^{-5}} = \boxed{71.73 C}[/tex]

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