Answer:
(a) 1.075 N
(b) 4.254 T
Explanation:
(a)
From the question,
Power = Force×Velocity.
P = Fv................... Equation 1
Where P = Power dissipated in the circuit, F = Pulling force of the wire, v = speed of the wire.
make F the subject of the equation
F = P/v................. Equation 2
Given: P = 4.30 W, v = 4.0 m/s.
Substitute into equation 2
F = 4.30/4
F = 1.075 N.
(b)
Applying,
F = BILsinФ.............. Equation 2
Where B = Strength of the magnetic field, I = current in the wire, L = Length of the wire, Ф = angle between the wire and the magnetic field.
make B the subject of the equation
B = F/ILsinФ................... Equation 3
But,
P = I²R...................... Equation 4
Where R = resistance of the wire.
make I the subject of the equation
I = √(P/R)............... Equation 5
Given: P = 4.30 W, R = 0.330 Ω
Substitute into equation 5
I = √(4.3/0.33)
I = √13.03
I = 3.61 A.
Also given: L = 7 cm = 0.07 m, Ф = 90°
Substitute into equation 3
B = 1.075/(0.07×3.61×sin90)
B = 1.075/0.2527
B = 4.254 T.
The pulling force and the strength of the magnetic field is mathematically given as
F=1.075N
I=3.61A
What are the pulling force and the strength of the magnetic field?
Question Parameters:
A 7.00-cm-long wire
The total resistance of the wire and rail is 0.330 Ω
Pulling the wire at a steady speed of 4.00 m/s causes 4.30 W of power.
Generally, the equation for the Power is mathematically given as
P=Fv
Therefore
F=P/v
F=4.30/4
F=1.075N
b)
Generally, the equation for the Current is mathematically given as
[tex]I=\sqrt{p/r}\\\\I=\sqrt{13.03}[/tex]
I=3.61A
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