Question

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

Answer 1

`4.77\ \text{A}`

Step-by-step explanation:

F = Magnetic force = 4.11 N

`I_n` = Net current

`I_2` = Current in one of the wires = 7.68 A

B = Magnetic field = 0.59 T

`\theta` = Angle between current and magnetic field =
`65^(\circ)`

`l` = Length of wires = 2.64 m

`I` = Current in the other wire

Magnetic force is given by

`F=I_nlB\sin\theta\\\Rightarrow I_n=(F)/(lB\sin\theta)\\\Rightarrow I_n=(4.11)/(2.64* 0.59 \sin65^(\circ))\\\Rightarrow I_n=2.91\ \text{A}`

Net current is given by

`I_n=I_2-I\\\Rightarrow I=I_2-I_n\\\Rightarrow I=7.68-2.91\\\Rightarrow I=4.77\ \text{A}`

The current I is
`4.77\ \text{A}`.