A 20.0 cm wire carrying a current of 10.0 A is placed in a uniform magnetic field of 0.30 T. If the wire makes an angle of 40° with the direction of the magnetic field, find the magnitude of the force acting on the wire? ( sin 40° = 0.642)
Correct answer: B. 0.39 N
- A. 2.71 N
- B. 0.39 N
- C. 6.61 N
- D. 7.61 N
- E. 9.91 N
Explanation
The force on a current-carrying conductor in a magnetic field is given by the formula: F = ILBsin(θ), where:I is the current (10.0 A)L is the length of the wire (0.20 m)B is the magnetic field strength (0.30 T)θ is the angle between the wire and the magnetic field (40°)Substituting the given values into the formula, we have:F = 10.0 A * 0.20 m * 0.30 T * sin(40°) = 0.39 NTherefore, the correct answer is 0.39 N. The other options are incorrect due to calculation errors, such as neglecting the sine function or using incorrect values for the variables.
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Magnetic flux density and magnetic flux describe the strength of a magnetic field and the field passing through a surface. A charged particle moving through a magnetic field experiences a force perpendicular to its velocity and may follow circular or helical motion, depending on the angle between velocity and field.
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