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Wize University Physics Textbook (Master) > Magnetic Fields and Magnetic Force

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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Magnetic Force on a Wire Element


Unfortunately, not every wire is a neat straight line! If the wire element is curved, we'll need to use calculus to find the magnetic force in some scenarios.
  • For a wire of arbitrary shape, the force on an infinitesimal piece of the wire, dl⃗d\vec{l}, is given by the following equation:
dF⃗=Idl⃗×B⃗d\vec{F}=Id\vec{l}\times\vec{B}
  • The total force is then found by integrating over the entire wire:
F⃗=∫dF⃗=∫Idl⃗×B⃗\vec F = \int d\vec F = \int I d\vec l \times \vec B
  • This can be used for curved wires, or cases where the magnetic field or current is not constant for all points on the wire.
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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 Wire (3D)


Calculate the force acting on a wire of length 2.0 meters which carries 8.0 A of current in the positive y-direction, through a magnetic field given by the following vector expression:

B⃗=2mTi^+3mTj^+4mTk^\vec{B}=2mT\hat{i}+3mT\hat{j}+4mT\hat{k}

To find the cross product, we will need to consider each component of the magnetic field individually. Remember the basic cross product results: i^×j^=k^;  j^×k^=i^;  k^×i^=j^;  i^×i^=0\hat i \times \hat j=\hat k; ~~\hat j \times \hat k= \hat i ; ~~ \hat k \times \hat i =\hat j; ~~\hat i \times \hat i =0.
F⃗=Il⃗×B⃗F⃗=(8A)(2m j^)×(0.002Ti^+0.003Tj^+0.004Tk^)F⃗=(8A)(−0.004mTk^+0.008mTi^)F⃗=0.064Ni^−0.032Nk^\begin{aligned} \vec{F}&=I\vec{l}\times\vec{B} \\ \vec{F}&=(8 A)(2m~ \hat{j})\times(0.002T\hat{i}+0.003T\hat{j}+0.004T\hat{k}) \\ \vec{F}&=(8A)(-0.004mT\hat{k}+0.008mT\hat{i}) \\ \vec{F}&=0.064N\hat{i}-0.032N \hat{k} \end{aligned}

Practice: Magnetic Force on a Square Wire


A square ring of current with side lengths a=15.0 cm lies in the x-y plane, centered at the origin. It is carrying a current I = 5.0 A, circulating counterclockwise when viewed from above the plane (i.e. from positive z). There is a uniform magnetic field of B⃗=5.0mTk^\vec{B}=5.0mT\hat{k} filling the region.


a) Calculate the net force on the right half of the square, where x > 0.
b) Calculate the net force on the left half of the square, where x < 0.
c) What is the net force on the ring?
Part a)

Practice: Magnetic Force on a Semi-Circular Wire


A circular half-ring of current with radius a=15.0 cm lies in the x-y plane, centered at the origin. The wire is carrying a current I = 5.0 A, circulating counterclockwise when viewed from above the plane (i.e. from positive z). There is a uniform magnetic field of B⃗=5.0mTk^\vec{B}=5.0mT\hat{k} filling the region.

Calculate the net force on this semicircular piece of current.