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Figure shows the to views of a rod that can slide without friction. The resistor is `6.0Omega` and a `2.5T` magnetic field is directed perpendicularly downward into the paper. Let `l=1.20m`. a. Calculate the force `F` required to move the rod to the right at a constant speed of `2.0 m/s`, b. At what rate is energy deliered to the resistor? c. Show that this rate is equal to the rate of work done by the applied force. |
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Answer» Given, resistance, R =`6.0Omega` Magnetic field, `B=2.57T` Length of the rod, `l=1.20m` The motional emf in the rod is given as, e = Bvl `or" "e=(2.5)(2.0)(1.2)V=6.0V` The current in the circuit, `i=(e)/(R)=(6.0)/(6.0)=1.0A` (i)The magnitude of force F required with be equal to the magnetic force acting on the rod, which oppses the motion. `therefore" "F=F_(m)=lvB=ilB` `or" "F=(1.0)(1.2)(2.5)N=3N` (ii) Rate by which energy is delivered to the resistor is `P_(1)=i^(2)R=(1)^(2)(6.0)=6W` (iii) The rate by which work is done by the applied force is, `P_(2)=Fv=(3)(2.0)=6WrArrP_(1)=P_(2)` |
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