A 0.289 m -long bar moves on parallel rails that are connected through a 5.56 ?

Question

A 0.289 m -long bar moves on parallel rails that are connected through a 5.56 ? resistor, as shown in the following figure (Figure 1) , so the apparatus makes a complete circuit. You can ignore the resistance of the bar and rails. The circuit is in a uniform magnetic field 1.35 Tthat is directed into the plane of the figure.

A) At an instant when the induced current in the circuit is counterclockwise and equal to 1.50 A , what is the magnitude of the velocity of the bar?

Express your answer with the appropriate units.

B) What is the direction of the velocity of the bar?

Explanation / Answer

Force in magnetic field is given as F_B = qvBsin(theta), theta is angle between field lines and bar

also F_B = i *L *B* sin (theta)

therefore v=(i * L)/q = (1.5 * 0.289)/1.6 x 10^-19

v = 2.71 x 10^18 Am C^-1 or 2.71 x 10^18 m s^-1

direction of the velocity of the bar is towards right

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