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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 in a uniform magnetic field of flux density 0.20 T0.20\text{ T}. If the wire experiences a magnetic force of 0.20 N0.20\text{ N}, what is the angle between the wire and the direction of the magnetic field?

  1. 30∘30^\circAnswer
  2. B
    45∘45^\circ
  3. C
    60∘60^\circ
  4. D
    90∘90^\circ

Answer

The angle between the wire and the direction of the magnetic field is 30∘30^\circ.
The magnetic force on a straight current-carrying wire in a uniform magnetic field is given by F=BILsin⁡θF = BIL\sin\theta. Substituting F=0.20 NF = 0.20\text{ N}, 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} gives 0.20=0.40sin⁡θ0.20 = 0.40 \sin\theta, which simplifies to sin⁡θ=0.50\sin\theta = 0.50. Therefore, θ=30∘\theta = 30^\circ.

Step-by-Step Solution

1
Identify the formula for the magnetic force on a current-carrying conductor in a magnetic field.
F=BILsin⁡θF = B I L \sin\theta
The magnetic force depends on magnetic flux density BB, current II, length LL, and the angle θ\theta between the conductor and the magnetic field.
2
Substitute the given values into the magnetic force formula.
0.20=0.20×4.0×0.50×sin⁡θ0.20 = 0.20 \times 4.0 \times 0.50 \times \sin\theta
Given values are F=0.20 NF = 0.20\text{ N}, 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}.
3
Solve for sin⁡θ\sin\theta and calculate θ\theta.
sin⁡θ=0.200.40=0.50  ⟹  θ=arcsin⁡(0.50)=30∘\sin\theta = \frac{0.20}{0.40} = 0.50 \implies \theta = \arcsin(0.50) = 30^\circ
Taking the inverse sine of 0.500.50 yields an angle of 30∘30^\circ.

Key Concept

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

Magnetic Force and Electromagnetism

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