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Wize High School Grade 12 Physics Textbook > Magnetism

Magnetic Force on Current-Carrying Wires

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Magnetic Force on a Wire


When we think about a wire of current, we're really just thinking about a lot of charges moving in the same way. As a result, the concepts of magnetic force on wires are very similar to the concepts for individual charges.
  • For an individual charge, the formula for magnetic force was F⃗=qv⃗×B⃗\vec F = q\vec v \times \vec B.
  • To find the magnitude, we can use ∣F⃗∣=qvBsin⁡θ|\vec F| = qvB\sin\theta
  • For a current of charge, the formulas are similar:
F⃗=Il⃗×B⃗∣F⃗∣=IlBsin⁡θ\boxed{\vec F = I \vec l \times \vec B} \\ \boxed{|\vec F|=IlB\sin\theta}
  • In this formula, the vector l⃗\vec l has magnitude equal to the length of the wire segment, and the direction is equal to the direction of positive current direction.


Wize Concept
If you have a closed loop of current in a uniform magnetic field, the total magnetic force on the wire will be zero.

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Example: Magnetic Force on a Wire


A wire of length 2.0 meters carries a current of 8.0 A flowing in the positive y-direction (up the page). This wire is placed in an external magnetic field of strength 2.0 mT pointing in the positive x-direction. Find the total magnetic force on the wire.


We will use the formula F⃗=Il⃗×B⃗\vec{F}=I\vec{l}\times\vec{B} to find the magnitude, with the direction being provided by the right-hand rule.

Magnitude:
∣F⃗∣=Il∣B⃗∣sin⁡θ∣F⃗∣=(8.0A)(2.0m)(0.002T)sin⁡(90o)∣F⃗∣=0.032 N\begin{aligned} |\vec{F}|&=Il|\vec{B}|\sin\theta \\ |\vec F| &= (8.0 A)(2.0m)(0.002T)\sin(90^o)\\ |\vec F| &= 0.032 ~N \end{aligned}
Direction:

Using the right-hand rule, point your right fingers up the page (direction of positive-y) and curl your fingers to the right (towards the direction of positive-x). You should find that your thumb points into the page, which is the negative-z direction.
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Example: Magnetic Force on a Long Current Carrying Wire

Calculate the force acting on a 2 m long wire which carries 8 A current in positive y-direction, through a magnetic field given byB⃗=(2mT)i^\vec{B}=\left(2mT\right)\hat{i} (the positive x direction).

F⃗=Il⃗×B⃗\vec{F}=I\vec{l}\times\vec{B}

F⃗=(8A)(+(2m)j^)×(2.0×10−3Ti^)\vec{F}=(8 A)(+(2m)\hat{j})\times(2.0\times 10^{-3}T \hat{i})

F⃗=0.032N(j^×i^)=−0.032Nk^\vec{F}=0.032 N (\hat{j}\times\hat{i})=-0.032 N\hat{k}