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Difficulty: HardMagnetic Force and Electromagnetism

Match each physical phenomenon or quantity involving electromagnetic forces listed on the left with its corresponding governing mathematical equation on the right.

  • Magnetic force exerted on a straight current-carrying conductor in a uniform magnetic fieldF=BILsin⁡θF = B I L \sin \theta
  • Magnetic force per unit length between two long parallel current-carrying conductors in vacuumFL=μ0I1I22πd\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}
  • Radius of the circular trajectory of a charged particle moving perpendicularly to a uniform magnetic fieldr=mvqBr = \frac{m v}{q B}
  • Torque experienced by a current-carrying rectangular coil suspended in a uniform magnetic fieldτ=BIANsin⁡θ\tau = B I A N \sin \theta

Answer

Magnetic force on a current-carrying conductor pairs with F=BILsin⁡θF = B I L \sin \theta; Force per unit length between parallel conductors pairs with FL=μ0I1I22πd\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}; Radius of circular trajectory of a charged particle pairs with r=mvqBr = \frac{m v}{q B}; Torque on a current-carrying coil pairs with τ=BIANsin⁡θ\tau = B I A N \sin \theta.
Each electromagnetic phenomenon matches directly with its derived expression from the magnetic force laws: magnetic force on a wire is F=BILsin⁡θF = B I L \sin \theta, force between parallel wires is FL=μ0I1I22πd\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}, orbit radius of charge is r=mvqBr = \frac{m v}{q B}, and coil torque is τ=BIANsin⁡θ\tau = B I A N \sin \theta.

Step-by-Step Solution

1
Analyze the magnetic force on a straight conductor
The Lorentz force law applied to current elements yields F=BILsin⁡θF = B I L \sin \theta.
Free charges moving inside the wire experience magnetic force perpendicular to both current and magnetic field vector.
2
Determine the mutual force formula for parallel conductors
The magnetic field from wire 1 is B1=μ0I12πdB_1 = \frac{\mu_0 I_1}{2\pi d}, giving force per length FL=B1I2=μ0I1I22πd\frac{F}{L} = B_1 I_2 = \frac{\mu_0 I_1 I_2}{2\pi d}.
Each conductor sits within the circular magnetic field lines generated by the other conductor.
3
Derive the motion equation for a charged particle in a magnetic field
Setting qvB=mv2rq v B = \frac{m v^2}{r} yields r=mvqBr = \frac{m v}{q B}.
The magnetic force provides the required inward centripetal acceleration for circular motion.
4
Identify the expression for torque on a magnetic dipole / coil
The couple produced by forces on opposite sides of a rectangular loop gives τ=BIANsin⁡θ\tau = B I A N \sin \theta.
Opposite sides experience forces in opposing directions separated by a moment arm.

Key Concept

Formulas for magnetic forces on current-carrying conductors, moving charges, parallel wires, and coils
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Topics

Magnetic Force and Electromagnetism

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