A conducting rod of mass m with negligible resistance rests on two horizontal fr

Question

A conducting rod of mass m with negligible resistance rests on two horizontal frictionless parallel rails separated by a distance l = 0.40 m as shown below. There is a constant magnetic field, B = 0.50 T (not shown), directed upward in the space between the rails. The left ends of the two rails are attached by a wire containing a resistor with resistance R = 50 ?. A varying external force, Fext(t), is applied to the rod causing it to oscillate back and forth with amplitude A = 0.20 m and angular frequency ? = 50 rad/s. The position of the rod as measured from its initial position (shown dotted) is x(t) and is given as a function of time.

(see attached)

Calculate the power dissipated in the resistor, PR = Ri^2 , and the power supplied by the external force, P_ext = F_ext*v. Show that the time average of these are equal compute their value. (Answer 40 mW)

5. A conducting rod of mass m with negligible resistance rests on two horizontal frictionless parallel rails shown), directed upward in the space between the rails. The left ends of the two rails are attached by a wire containing a resistor with resistance R-50 . A varying external force, Fet(t), is applied to the rod causing it to oscillate back and forth with amplitude A 0.20 m and angular frequency w 50 rad/s. The position of the rod as measured from its initial position (shown dotted) is (t) and is given as a function of time. x(t) )Asinut) A sin(wt) Calculate the power dissipated in the resistor Pa = R, and the power supplied by the external force Pext Fext. Show that the time average of these are equal compute their value. (Answer 40 mW)

Explanation / Answer

Here,    emf induced in rod = Blv

                                           = B * l * dx/dt

                                           = 0.50 * 0.40 * Aw * cos(wt)

                                           = 0.50 * 0.40 * 0.20 * 50 * cos(50t)

                                           =   2 * cos(50t)

=>    power dissipated in the resistor   =   (2 * cos(50t))2/50

                                                              =   0.08 * cos2(50t)

=>     power supplied by the external force   = (I * L * B) * v

                                                                       = (2 * cos(50t)/50) * 0.40 * 0.50 * 0.20 * 50 * cos(50t)

                                                                       =   0.08 * cos2(50t)

=>   time average    =   0.08 * (1/2) =    0.04 W

                                                 =   40 mW

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