Answer
given,
length of wire = 10 cm = 0.1 m
resistance of the wire = 0.330 ohms
speed of pulling = 4 m/s
Power = 4.40 W
a) Force of pull = ?
[tex]P = F_{pull} v[/tex]
[tex]F_{pull} =\dfrac{P}{v}[/tex]
[tex]F_{pull} =\dfrac{4.40}{4}[/tex]
[tex]F_{pull} =1.1\ N[/tex]
b) using formula
[tex]P = \dfrac{B^2l^2v^2}{R}[/tex]
where B is the magnetic field
v is the pulling velocity
R is the resistance of the wire
[tex]B =\sqrt{\dfrac{PR}{l^2v^2}}[/tex]
[tex]B =\sqrt{\dfrac{F_{pull} \times v R}{l^2v^2}}[/tex]
[tex]B =\sqrt{\dfrac{F_{pull} \times R}{l^2v}}[/tex]
[tex]B =\sqrt{\dfrac{1.1 \times 0.33}{0.1^2\times 4}}[/tex]
[tex]B =\sqrt{9.075}[/tex]
B = 3.01 T
The pulling force in this magnetic field is equal to 1.1 Newton.
Given the following data:
Mathematically, the pulling force in a magnetic field is given by this formula:
[tex]F = \frac{Power}{Speed} \\\\F=\frac{4.40}{4.00}[/tex]
F = 1.1 Newton.
To determine the strength of the magnetic field, we would apply this formula:
[tex]B=\sqrt{\frac{PR}{L^2V^2} }\\\\B=\sqrt{\frac{FVR}{L^2V^2} }\\\\B=\sqrt{\frac{FR}{L^2V} }[/tex]
Where:
Substituting the given parameters into the formula, we have;
[tex]B=\sqrt{\frac{1.1 \times 0.330}{0.1^2 \times 4.00} } \\\\B=\sqrt{\frac{0.363}{0.01 \times 4.00} }\\\\B=\sqrt{9.075}[/tex]
B = 3.012 T.
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