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]
