A straight wire 15cm long, carrying a current of 6.0A is in a uniform field of 0.40T. What is the force on the wire when it is at right angle to the field

The force on a current-carrying wire in a uniform magnetic field can be calculated using the equation: F = BILsinθ where F is the force in Newtons, B is the magnetic field strength in Tesla, I is the current in Amperes, L is the length of the wire in meters, and θ is the angle between the wire and the magnetic field. In this problem, the wire is 15cm long (0.15m), carrying a current of 6.0A, and the magnetic field is 0.40T. The angle between the wire and the magnetic field is 90 degrees (since the wire is at right angles to the field). Substituting the given values into the equation, we get: F = (0.40T)(6.0A)(0.15m)sin90 sin90 = 1, so we can simplify the equation to: F = (0.40T)(6.0A)(0.15m) F = 0.36N Therefore, the force on the wire is 0.36N. Answer option C is the correct answer.