A) 1.05 N
The power dissipated in the circuit can be written as the product between the pulling force and the speed of the wire:
[tex]P=Fv[/tex]
where
P = 4.20 W is the power
F is the magnitude of the pulling force
v = 4.0 m/s is the speed of the wire
Solving the equation for F, we find
[tex]F=\frac{P}{v}=\frac{4.20 W}{4.0 m/s}=1.05 N[/tex]
B) 3.03 T
The electromotive force induced in the circuit is:
[tex]\epsilon=BvL[/tex] (1)
where
B is the strength of the magnetic field
v = 4.0 m/s is the speed of the wire
L = 10.0 cm = 0.10 m is the length of the wire
We also know that the power dissipated is
[tex]P=\frac{\epsilon^2}{R}[/tex] (2)
where
[tex]R=0.350 \Omega[/tex] is the resistance of the wire
Subsituting (1) into (2), we get
[tex]P=\frac{B^2 v^2 L^2}{R}[/tex]
And solving it for B, we find the strength of the magnetic field:
[tex]B=\frac{\sqrt{PR}}{vL}=\frac{\sqrt{(4.20 W)(0.350 \Omega)}}{(4.0 m/s)(0.10 m)}=3.03 T[/tex]
