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The magnetic field B at all points within the colored circle of the figure (Figure 1) has an initial magnitude of 0.780 T. (The circle could represent approximately the space inside a long, thin solenoid.) The magnetic field is directed into the plane of the diagram and is decreasing at the rate of 0.0300 T/s.
a) What is the magnitude of the induced electric field at any point on the circular conducting ring with radius r = 0.100 m ?
b) What is the direction of this field at any point on the circular conducting ring?
c) What is the current in the ring if its resistance is 4.00 Ω ?
d) What is the emf between points a and b on the ring?
e) If the ring is cut at some point and the ends are separated slightly, what will be the emf between the ends?

The magnetic field B at all points within the colored circle of the figure Figure 1 has an initial magnitude of 0780 T The circle could represent approximately class=

Respuesta :

The magnitude of the induced electrical field is 0.0015V/m, the field is pointing towards the clockwise direction while the current in the ring will be 0.0002355A if the resistance is 4 ohms. The emf between point a and b is zero and the EMF across the point if they're slightly separated between the ends is 0.000942V

To solve this question, we would have to go about each one individually

Data:

[tex]r=10cm=0.1m\\[/tex]

a.

The magnitude of the induced electrical field at any point within the radius is

[tex]\int\limits^a_b {E} \, du=\frac{dU}{dt}=\pi \frac{dB}{dt}=\pi r^{2}\frac{dB}{dt}\\E*2\pi r=\pi r^{2}\frac{dB}{dt} \\E=\frac{r}{2}\frac{dB}{dt}=\frac{0.1}{2}*0.03=0.0015V/m[/tex]

b.

The field is pointing towards the clockwise direction.

c.

The current in the ring if we are given a resistance of 4ohms

[tex]I=\frac{emf}{R}=\frac{\pi r^{2}\frac{dB}{dt} }{R} =\frac{\pi (0.1)^2*0.03}{4} =0.0002355A[/tex]

d.

The emf between point a and b is zero

e.

The EMF across two points if they're separated by small distance across the ring is

we would use the formula to solve for the EMF

[tex]E=\pi r^{2}\frac{dB}{dt}=\pi (0.1)^2*0.03=0.000942V[/tex]

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