contestada

The loop is in a magnetic field 0.32 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A = 0.285 m2.Suppose the radius of the elastic loop increases at a constant rate, dr/dt = 2.70 cm/s .
1) Determine the emf induced in the loop at t = 0
2) Determine the emf induced in the loop at t = 1.00 s .

Respuesta :

Answer:

(a) - [tex]emf=0.0163 \ V}}[/tex]

(b) - [tex]emf=0.0178 \ V}}[/tex]

Explanation:

Induced emf (or voltage) can be calculated using the following formula.

[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Induced Emf:}}\\\\||emf||=N\frac{d\Phi_b}{dt} \end{array}\right}[/tex]

Where...

"N" represents the number of turns/coils of wire

"dΦ_B" represents the change in magnetic flux

"dt" represents the change in time

In this case N=1, so we have the equation...

[tex]emf=\frac{d\Phi_b}{dt}[/tex]

Magnetic flux can be calculated as follows.

[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Magnetic Flux:}}\\\\ \Phi_b=BA\cos(\theta) \end{array}\right}[/tex]

Where...

"B" represents the strength of the magnetic field

"A" represents the area of a surface

"θ" represents the angle between B and A

In this case θ=0°, so we have the equation..

[tex]\Phi_B=BA[/tex]

Given:

[tex]B=0.32 \ T\\A_0=0.285 \ m^2\\\frac{dr}{dt}=2.70 \ cm/s \rightarrow 0.027 \ m/s[/tex]

Find:

[tex]emf \ \text{when} \ dt=0 \ s \\\\emf \ \text{when} \ dt=1.00 \ s[/tex]

(1) - Find the initial radius of the loop

[tex]\text{Recall the area of a circle} \rightarrow A=\pi r^2\\\\A_0=\pi r_0^2\\\\\Longrightarrow r_0=\sqrt{\frac{A_0}{\pi} } \\\\\Longrightarrow r_0=\sqrt{\frac{0.285}{\pi} } \\\\\therefore \boxed{r_0 \approx 0.301 \ m}[/tex]

(2) - Find dΦ_B/dt

[tex]\Phi_B=BA\\\\\Longrightarrow \Phi_B=B(\pi r^2)\\\\\Longrightarrow \frac{d\Phi_B}{dt} =B( 2\pi r)\frac{dr}{dt} \\\\\therefore \boxed{emf=2B\pi r\frac{dr}{dt}}[/tex]

(3) - For part (a) plug in the appropriate values into the equation

[tex]emf=2B\pi r\frac{dr}{dt}\\\\\Longrightarrow emf=2(0.32)(\pi)(0.301)(0.027)\\\\\therefore \boxed{\boxed{emf=0.0163 \ V}}[/tex]

(4) - Find the radius of the loop after one second

[tex]r_f=r_0+\frac{dr}{dt} \\\\\Longrightarrow r_f=0.301+0.027\\\\\therefore \boxed{r_f=0.328}[/tex]

(5) - Use the new radius value to answer part (b)

[tex]emf=2B\pi r\frac{dr}{dt}\\\\\Longrightarrow emf=2(0.32)(\pi)(0.328)(0.027)\\\\\therefore \boxed{\boxed{emf=0.0178 \ V}}[/tex]

Thus, the problem is solved.