The following figure shows a uniform magnetic field B into...

The following figure shows a uniform magnetic field B into the page confined to a cylindrical volume of radius R. The magnitude of B is decreasing at a constant rate of 10. mT/s. What is the initial acceleration of an electron released at point a (at coordinates (x, y) = (0, -5.0cm))? A) 250 µm/s² to the left B) 250 µm/s² to the right C) 250 µm/s² downwards D) 250 µm/s² out of the page E) 250 µm/s² into the page F) 4.4 x 10$^7$ m/s² to the left G) 4.4 x 10$^7$ m/s² to the right H) 4.4 x 10$^7$ m/s² downwards I) 4.4 x 10$^7$ m/s² out of the page J) 4.4 x 10$^7$ m/s² into the page

Transcript

00:01 In this problem on the topic of induction, we are shown in the figure a uniform magnetic field b that is confined to a cylindrical volume which has a radius r.

00:08 The magnitude of the field is decreasing at a constant rate of 10 millitres per second, and we want to find the initial acceleration of an electron that is released at point a, which is at a radial distance of 5 centimeters, a point b at r is equal to 0, and c at r is equal to 5 centimeters.

00:27 Now the changing magnetic field induces an electric field, and the induced electric field is given by the closed integral of e.

00:42 Ds is equal to minus the rate of change of magnetic flux, d5b, d t .t.

00:52 The electric field lines are circles that are concentric with the cylindrical region, and so therefore we have e.

01:01 Times 2 pi r is equal to minus pi r squared d b d t which means that the electric field magnitude e is minus a half d b d t times r and the force on the electron f is equal to minus the charge of the electron e times the field vector e times the field vector big e, and newton's second law therefore tells us that the acceleration on the electron a is equal to this force minus little e times big e over the electron's mass, m.

01:51 So firstly, at point a, we have the electric field strength e to be minus r over 2 db, d t.

02:08 And so this is, and so this is, minus a half times 5 times 10 to the minus 2 meters times 10 to the minus 10 times 10 to the minus 3 tesla's per second which is dbdt and so we get the electric field strength at point a to be 2 .5 times 10 to the minus 4 volts per meter and so actually we want the acceleration here...

Question

Please give Ace some feedback