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Two insulated wires, each 2.64 m long, are taped together to form a two-wire unit that is 2.64 m long. One wire carries a current of 7.68 A; the other carries a smaller current I in the opposite direction. The two wire unit is placed at an angle of 65.0o relative to a magnetic field whose magnitude is 0.59 T. The magnitude of the net magnetic force experienced by the two-wire unit is 4.11 N. What is the current I

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Answer:

[tex]4.77\ \text{A}[/tex]

Explanation:

F = Magnetic force = 4.11 N

[tex]I_n[/tex] = Net current

[tex]I_2[/tex] = Current in one of the wires = 7.68 A

B = Magnetic field = 0.59 T

[tex]\theta[/tex] = Angle between current and magnetic field = [tex]65^{\circ}[/tex]

[tex]l[/tex] = Length of wires = 2.64 m

[tex]I[/tex] = Current in the other wire

Magnetic force is given by

[tex]F=I_nlB\sin\theta\\\Rightarrow I_n=\dfrac{F}{lB\sin\theta}\\\Rightarrow I_n=\dfrac{4.11}{2.64\times 0.59 \sin65^{\circ}}\\\Rightarrow I_n=2.91\ \text{A}[/tex]

Net current is given by

[tex]I_n=I_2-I\\\Rightarrow I=I_2-I_n\\\Rightarrow I=7.68-2.91\\\Rightarrow I=4.77\ \text{A}[/tex]

The current I is [tex]4.77\ \text{A}[/tex].

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