A conducting rod AB of length d = 0.90 m and mass 0.30 kg makes contact with the

Question

A conducting rod AB of length d = 0.90 m and mass 0.30 kg makes contact with the metal rails AD and BC as shown in the diagram. The apparatus is in a uniform magnetic field of flux density 590 mWb/m2 (1 Wb = 1 Tm2), perpendicular to the plane of the diagram. At the moment shown, distance AD is 2.30 m. Find the magnitude of the induced emf in the rod if it is moving to the right with a velocity of 6.60 m/s. Correct, computer gets: 3.50E+00 V Hint: Use self-consistent units! "flux density" is a fancy word for "magnetic field" (look at the units, 1 Wb/m^2 is the same as Teslas) Which of the given quantities do you need to solve this problem? 4.[1pt] Answer the following questions with regard to the diagram above. If the first is T and the rest F, enter TFFFF. A) In the absence of friction between rod and rail, the rod will continue to move towards the right indefinitely even if no external force is applied to keep it in motion. B) The induced current flows clockwise. C) As the rod AB moves to the right, the magnitude of the flux through the surface ABCD decreases. D) The agent moving the rod to the right must be doing zero work on the rod E) The magnetic field due to the INDUCED current points within the loop ABCD out of the plane of the paper. Correct, computer gets: FTTFF Hint: This is an application of Faraday's and Lenz's Laws. 5.[1pt] In the problem above, the magnetic field was normal, i.e. it made an angle of 90 degrees with the plane of the loop ABCD. Now suppose the magnetic field makes an angle of 40o with the plane of the loop ABCD. Find the induced emf. (Assume the same flux density and velocity given before.)

Explanation / Answer

1. B = 590 mW/m² , d = 0.9m, v = 6.60 m/sec.
Induced emf = / t = B * d*x/ t = B*d*v = 0.59*0.9*6.6 = 3.505 V

5. When the magnetic field makes an angle of 40° with the plane of the loop , the normal component to the plane is B*sin 40 . As the loop and rod geometry is not changed

the induced emf = B*sin40*d*v = 0.59*sin 40*0.9*6.6 = 0.912*sin 40 = 2.611 V

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