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Zorluk: Çok zorMagnetic Force and Electromagnetism

A straight horizontal wire of length 0.40 m0.40\text{ m} and mass 0.12 kg0.12\text{ kg} carries a steady electric current directed towards the East. The wire is situated in a uniform horizontal magnetic field of 0.60 T0.60\text{ T} directed at an angle of 30∘30^\circ North of East. Taking the acceleration due to gravity g=10 m s−2g = 10\text{ m s}^{-2}, what minimum current must flow through the wire for the magnetic force to act vertically upward and balance the weight of the wire?

  1. A
    5.0 A5.0\text{ A}
  2. 10.0 A10.0\text{ A}Cevap
  3. C
    1.0 A1.0\text{ A}
  4. D
    5.8 A5.8\text{ A}

Cevap

10.0 A10.0\text{ A}
The magnetic force on a current-carrying conductor is given by F=ILBsin⁡θF = I L B \sin\theta. With current flowing East and the magnetic field directed 30∘30^\circ North of East, the angle θ=30∘\theta = 30^\circ. The right-hand rule confirms that L⃗×B⃗\vec{L} \times \vec{B} points vertically upward. Setting the upward force equal to the weight mgmg, we get I×0.40×0.60×sin⁡30∘=0.12×10I \times 0.40 \times 0.60 \times \sin 30^\circ = 0.12 \times 10, which yields I=10.0 AI = 10.0\text{ A}.

Adım Adım Çözüm

1
Calculate the weight of the horizontal wire
W=mg=0.12 kg×10 m s−2=1.2 NW = mg = 0.12\text{ kg} \times 10\text{ m s}^{-2} = 1.2\text{ N}
The magnetic force must balance the downward gravitational force acting on the wire.
2
Determine the angle θ\theta between the current direction and magnetic field
θ=30∘\theta = 30^\circ
The current flows East and the magnetic field points 30∘30^\circ North of East, so the angle between the vector length element and magnetic field is 30∘30^\circ.
3
Express the magnetic force using F=ILBsin⁡θF = I L B \sin\theta
F=I×0.40 m×0.60 T×sin⁡30∘=0.12IF = I \times 0.40\text{ m} \times 0.60\text{ T} \times \sin 30^\circ = 0.12 I
The cross product IL⃗×B⃗\vec{I L} \times \vec{B} gives the magnitude ILBsin⁡θI L B \sin\theta and a vertical upward direction according to the right-hand rule.
4
Equate the magnetic force to the weight and solve for II
0.12I=1.2  ⟹  I=10.0 A0.12 I = 1.2 \implies I = 10.0\text{ A}
For complete vertical equilibrium, the upward magnetic force must equal the downward weight.

Anahtar Kavram

Magnetic force on a current-carrying conductor in a uniform magnetic field
Tahmini Süre:2m 0s
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Konular

Magnetic Force and Electromagnetism

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