Option (d) is correct.
The speed of the rod is 6.5 m/s
It is given that a uniform magnetic field of magnitude B = 3.0 T is perpendicular to the plane formed by the rod and the conducting rails.
Length of the rod l = 1m
resistance of the rod, R = 5Ω
mass of the rod m = 1.2 kg
Since the rod is sliding, the area enclosed by the rod and the rails is changing, so the magnetic flux through it is changing. Therefore, an EMF will be generated, given by:
[tex]E = Blv[/tex]
where v is the velocity of the rod.
Now a current will be generated in the rod due to this EMF given by:
[tex]I = E/R\\\\I =Blv/R[/tex]
The magnetic force on the current-carrying rod is given by:
[tex]F=BIl[/tex]
[tex]F=\frac{B^2l^2v}{R}[/tex]
But to keep the in equilibrium and sliding with constant velocity on the rails, the magnetic force must balance the weight of the rod:
[tex]F=\frac{B^2l^2v}{R}=mg\\\\v=\frac{mgR}{B^2l^2}\\\\v=\frac{1.2\times9.8\times5}{3^2\times1^2}\\\\v=6.53\;m/s[/tex]
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