Respuesta :
Question:
What are the magnitudes of the pulling force and the magnetic field
Answer:
The magnetic field strength is 2.7 T
The pulling force [tex]F_p[/tex] is 0.975 N
Explanation:
Here we have;
Length of wire, l = 10.0 cm = 0.1 m
Resistance of wire, R = 0.300 Ω
Speed of wire, v = 4.00 m/s
Power dissipated = 3.90 W
Based on the given data, we apply the relation;
[tex]I =\frac{Blv}{R}[/tex]..............(1)
Where:
B = Magnetic field strength
I = Current
Since P = I²R, we have;
3.9 = I²·0.300Ω
∴ I² = 3.9/0.300 = 13
From which I = √13
Substituting the value of I in equation (1) above, we have;
[tex]\sqrt{13}= \frac{B \times 10.0 \times 4.00}{0.300}[/tex]
Therefore;
[tex]B = \frac{\sqrt{13} \times 0.300}{0.1 \times 4.00} = 2.70416 \ T\approx 2.7 \ T[/tex]
The magnitude of the pulling force is given by the following relation;
[tex]F_p = F_m = IlB = l\sqrt{\frac{P}{R} } \frac{\sqrt{PR} }{lv} =\frac{P}{v} = \frac{3.9 \ W}{4.00 \ m/s} = 0.975 \ N[/tex]
The pulling force [tex]F_p[/tex] = 0.975 N.
