Two insulated wires, each 2.70 m long, are taped together to form a two-wire unit that is 2.70 m long. One wire carries a current of 7.00 A; the other carries a smaller current I in the opposite direction. The two wire unit is placed at an angle of 77.0° relative to a magnetic field whose magnitude is 0.360 T. The magnitude of the net magnetic force experienced by the two-wire unit is 3.13 N. What is the current I?
To find the current I, we can use the formula for the magnetic force experienced by a current-carrying wire in a magnetic field:
F = I * L * B * sin(theta),
where F is the net magnetic force, I is the current, L is the length of the wire, B is the magnetic field strength, and theta is the angle between the wire and the magnetic field.
In this case, the net magnetic force is given as 3.13 N, the length (L) of each wire is 2.70 m, the magnetic field strength (B) is 0.360 T, and theta is the angle between the wire unit and the magnetic field, which is 77.0°.
Now, let's rearrange the formula to solve for the current I:
I = F / (L * B * sin(theta)).
Substituting the given values into the formula:
I = 3.13 N / (2 * 2.70 m * 0.360 T * sin(77.0°)).
Now, we can calculate the value of I:
I = 3.13 N / (2 * 2.70 m * 0.360 T * sin(77.0°)).
I ≈ 0.423 A.
Therefore, the current I in the smaller wire is approximately 0.423 A.