The work done by the rod to overcome ampere force is: wa = magdsin53 = 2.4j. 。
Same reason? Wb=mbgdsin53 =0.8J
(2) In the process of A passing through the magnetic field, A is the power supply, B and R are external circuits, and Ia=Ib+IR.
IbRb=IRR，Ib=IR3，
So IAIB = 4, then qaqb = 163.
(3) Let the speed of B moving at a uniform speed in the magnetic field be vb, then the current IB in B = blvbr is 1.
The total resistance r of the circuit is1=14Ω.
According to the force balance, B2L2vbR is 1 = MBG SIN 53.
Similarly, when rod A moves at a uniform speed in a magnetic field, R is always 2 = 7 Ω.
B2L2vaR total 2 = magsin53。
Available va: VB = 3: 2
(4) From the meaning of the question: VA = VB+GSIN53 T, d=vbt.
Because v2 = v2=2gsin53 S.
The distances from point M and point N to L 1 are Sa=98m and Sb= 12m, respectively.
SMN = 5800 million.