In the problem above, the magnetic field was normal, i.e. it made an angle of 90

Question

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 50o with the plane of the loop ABCD. Find the induced emf. (Assume the same flux density and velocity given before.)

A conducting rod AB of length d -0.30 m and mass 0.20 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 680 mWb/m2 (1 Wb - 1 Tm2), perpendicular to the plane of the diagram. At the moment shown, distance AD is 2.10 m. Find the magnitude of the induced emf in the rod if it is moving to the right with a velocity of 1.60 m/s 21 Correct, computer gets: 3.26E-01 V

Explanation / Answer

When the magnetic field makes an angle of 30 with the plane of the loop , the normal component to the plane is B*sin 30 . As the loop and rod geometry is not changed

the induced emf = B*sin30*d*v = 0.38*sin 30*1.5*1.6 = 0.912*sin 30 = 0.456 V

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