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Difficulty: MediumMagnetic Force and Electromagnetism

A straight conductor of length 0.5 m0.5\text{ m} carries a current of 4.0 A4.0\text{ A} in a uniform magnetic field of flux density 0.20 T0.20\text{ T}. If the conductor is placed at an angle of 30∘30^\circ to the magnetic field lines, what is the magnitude of the magnetic force exerted on the conductor?

  1. 0.20 N0.20\text{ N}Answer
  2. B
    0.35 N0.35\text{ N}
  3. C
    0.40 N0.40\text{ N}
  4. D
    0.80 N0.80\text{ N}

Answer

0.20 N0.20\text{ N}
The magnetic force on a straight wire carrying current in a uniform magnetic field is given by F=BILsin⁡θF = BIL \sin\theta. Substituting the given values B=0.20 TB = 0.20\text{ T}, I=4.0 AI = 4.0\text{ A}, L=0.5 mL = 0.5\text{ m}, and θ=30∘\theta = 30^\circ yields F=0.20×4.0×0.5×0.5=0.20 NF = 0.20 \times 4.0 \times 0.5 \times 0.5 = 0.20\text{ N}.

Step-by-Step Solution

1
Identify the given physical quantities
L=0.5 mL = 0.5\text{ m}, I=4.0 AI = 4.0\text{ A}, B=0.20 TB = 0.20\text{ T}, and θ=30∘\theta = 30^\circ.
Extracting values from the problem statement ensures correct substitution.
2
State the magnetic force equation for a straight current-carrying conductor in a magnetic field
F=BILsin⁡θF = BIL \sin\theta
Only the component of the magnetic field perpendicular to the current contributes to the magnetic force.
3
Substitute the values and calculate the force magnitude
F=(0.20 T)×(4.0 A)×(0.5 m)×sin⁡(30∘)=0.20×4.0×0.5×0.5=0.20 NF = (0.20\text{ T}) \times (4.0\text{ A}) \times (0.5\text{ m}) \times \sin(30^\circ) = 0.20 \times 4.0 \times 0.5 \times 0.5 = 0.20\text{ N}.
Evaluating sin⁡30∘=0.5\sin 30^\circ = 0.5 gives the required force value.

Key Concept

Magnetic Force on a Current-Carrying Conductor
Estimated Time:1m 15s
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Topics

Magnetic Force and Electromagnetism

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