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Difficulty: EasyMagnetic Force and Electromagnetism

A straight wire of length 0.50 m0.50\text{ m} carrying a current of 4.0 A4.0\text{ A} is placed perpendicular to a uniform magnetic field of flux density 0.20 T0.20\text{ T}. What is the magnitude of the magnetic force acting on the wire?

  1. 0.40 N0.40\text{ N}Answer
  2. B
    0.80 N0.80\text{ N}
  3. C
    1.00 N1.00\text{ N}
  4. D
    0.00 N0.00\text{ N}

Answer

The magnetic force acting on the wire is 0.40 N0.40\text{ N}.
Using the formula F=BILsin⁡θF = BIL \sin\theta, when a current-carrying wire is perpendicular to the magnetic field, θ=90∘\theta = 90^\circ and sin⁡90∘=1\sin 90^\circ = 1. Substituting B=0.20 TB = 0.20\text{ T}, I=4.0 AI = 4.0\text{ A}, and L=0.50 mL = 0.50\text{ m} yields F=0.20×4.0×0.50=0.40 NF = 0.20 \times 4.0 \times 0.50 = 0.40\text{ N}.

Step-by-Step Solution

1
Identify the given values and formula
L=0.50 mL = 0.50\text{ m}, I=4.0 AI = 4.0\text{ A}, B=0.20 TB = 0.20\text{ T}, θ=90∘\theta = 90^\circ, and F=BILsin⁡θF = B I L \sin\theta.
The magnetic force on a current-carrying conductor in a uniform magnetic field depends on field strength, current, length, and angle of orientation.
2
Substitute the values into the formula and calculate
F=0.20×4.0×0.50×sin⁡(90∘)=0.40 NF = 0.20 \times 4.0 \times 0.50 \times \sin(90^\circ) = 0.40\text{ N}.
Since sin⁡90∘=1\sin 90^\circ = 1, the force is maximum for a given magnetic field and current.

Key Concept

Magnetic force on a current-carrying conductor (F=BILsin⁡θF = BIL \sin\theta)
Estimated Time:45s
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Topics

Magnetic Force and Electromagnetism

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