A 0.18 m section of wire is placed in a uniform magnetic field of 0.40 T. The wire carries a current of 1.70 A. The angle between the magnetic field and the wire is [math: 67.0∘]. What is the magnitude of the force experienced by the wire?

Question

BlackTom AI · Accepted Answer

The magnetic force on a current-carrying wire is F = I L B sinθ, so this formula and the sine of the angle are the key concepts.

1) Relevant concepts
- Force on a straight wire in a uniform magnetic field: F = I L B sinθ, where I is current, L is length, B is magnetic field magnitude, and θ is the angle between the wire and the magnetic field.
- Use the sine because only the component of the wire perpendicular to B produces force.

2) Step-by-step reasoning
- Plug the given values into the formula: I = 1.70 A, L = 0.18 m, B = 0.40 T, θ = 67.0°.
- Compute the product I·L first, then multiply by B, then multiply by sinθ.

3) Calculations (all intermediate steps)
- I·L = 1.70 A × 0.18 m = 0.306 A·m
- (I·L)·B = 0.306 × 0.40 T = 0.1224 N (this would be the force if θ = 90°)
- sin 67.0° ≈ 0.9205
- F = 0.1224 × 0.9205 ≈ 0.1126 N
- Rounding to two significant figures appropriate for the given data gives F ≈ 0.11 N

4) Why this answer is correct
- The derivation follows directly from the vector cross product magnitude F = I (L × B) = I L B sinθ. Using the correct trigonometric factor for the specified 67.0° angle yields ≈0.1126 N, which rounds to 0.11 N. This matches the physical expectation that the force is less than the maximum (when θ = 90°) because sin67° < 1.

5) Why the other options are incorrect
- 0.12 N: slightly higher; corresponds to not applying the sin factor correctly or rounding 0.1224 → 0.12 instead of including sin67°.  
- 0.05 N: about 0.0478 N if one used cos67° (wrong trig function), so this comes from using cos instead of sin.  
- 0.29 N: much larger than any correct product of the given values; inconsistent with the inputs.

6) Conclusion
The computed magnitude of the force is approximately 0.1126 N, which rounds to 0.11 N. This confirms the correct choice: 0.11 N.

A 0.18 m section of wire is placed in a uniform magnetic field of 0.40 T. The wire carries a current of 1.70 A. The angle between the magnetic field and the wire is [math: 67.0∘]. What is the magnitude of the force experienced by the wire?单项选择题

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