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

The Biot-Savart Law

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Biot-Savart Law for a Point Charge


A moving charge generates a magnetic field. The purpose of the Biot-Savart Law is to determine the exact magnetic field that is generated by moving charges.
  • Consider a point chargeqqmoving with velocityv⃗\vec{v}. If we consider a point in space that is separated from the point charge by r⃗\vec r, the magnetic field produced at this point is given by either of the following formulas:
B⃗=μ04πqv⃗×r^∣r⃗∣2∣B⃗∣=μ04πq∣v⃗∣sin⁡θ∣r⃗∣2\boxed{\vec{B}=\frac{\mu_0}{4\pi}\frac{q\vec{v}\times\hat{r}}{\left|\vec{r}\right|^2}}\\ \boxed{|\vec B|=\frac{\mu_0}{4\pi}\frac{q|\vec{v}|\sin\theta}{\left|\vec{r}\right|^2}}

Wize Tip
You might also encounter the formula B⃗=μ04πqv⃗×r⃗∣r⃗∣3\vec{B}=\frac{\mu_0}{4\pi}\frac{q\vec{v}\times\vec{r}}{\left|\vec{r}\right|^3}. This version of the Biot-Savart Law is identical to the one shown above. r⃗\vec r is a position vector with some magnitude and direction. Alternatively, r^\hat r is a unit vector (magnitude of one) with the same direction as r⃗\vec r.
  • In these formulas we have introduced the permeability of free space, μo=4π×10−7m kgs2A2\mu_o=4\pi\times10^{-7} \frac{m~kg}{s^2A^2}.
  • The angle θ\theta is the angle between the position vector and the velocity vector.
  • The magnetic field lines around a moving charged particle form closed circles.

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Example: Biot-Savart for a Point Charge

As seen in the following diagram, two positively charged particles of charge q move towards the origin. One charge travels in the positive-x direction at a constant speed of v, and the other travels in the positive-y direction at a constant speed of 2v.

a) What is the net magnetic field at the origin if both particles are a distance d away from the origin? (Neglect the interactions between the two charges.)

b) What is magnitude of the magnetic field halfway between the two particles at the same moment?

Part a)


The magnetic field produced by a moving charge depends on the angle between the velocity vector and the position vector (see the Biot-Savart law formula). For both charges, the charge is moving directly towards the origin, so the vectors are parallel, giving a magnetic field of zero.

Part b)


Let's refer to the point in question as "A".

By the Pythagorean theorem, we can tell that there is a distance of d2d\sqrt 2 between the two charges. Therefore, point A is at half of this distance from each charge - that is, ∣r⃗∣=22d|\vec r|=\frac{\sqrt2}{2}d. We can also tell that the angle between the position and velocity vectors is 45o45^o in each case.

Let's first find the magnetic field contribution from the charge moving in the horizontal direction. Plugging the values into the Biot-Savart law:
∣B⃗∣=μo4πq∣v⃗∣sin⁡θ∣r⃗∣2∣B⃗∣=μo4πqvsin⁡(45o)(22d)2∣B⃗∣=μo4πqv(22)(22)2(d)2∣B⃗∣=μo4πqv(22)d2\begin{aligned} |\vec B|&=\frac{\mu_o}{4 \pi} \frac{q|\vec v|\sin\theta}{|\vec r|^2} \\ |\vec B|&=\frac{\mu_o}{4 \pi} \frac{qv\sin(45^o)}{(\frac{\sqrt{2}}{2}d)^2}\\ |\vec B|&=\frac{\mu_o}{4 \pi} \frac{qv(\frac{\sqrt{2}}{2})}{(\frac{\sqrt{2}}{2})^2(d)^2}\\ |\vec B|&=\frac{\mu_o}{4 \pi} \frac{qv}{(\frac{\sqrt{2}}{2})d^2}\\ \end{aligned}
To find the direction, we take the cross product v⃗×r^\vec v \times \hat r. That is, we point our right-hand fingers along the velocity vector (to the right) and curl them in the direction of the position vector (down the page). Our thumb now points into the page which is the direction of this field contribution. (We keep this result because the charge is positive.)

Repeating this for point B, which is moving twice as fast:
∣B⃗∣=μo4πq∣v⃗∣sin⁡θ∣r⃗∣2∣B⃗∣=μo4πq(2v)sin⁡(45o)(22d)2∣B⃗∣=μo4π2qv(22)(22)2(d)2∣B⃗∣=μo4π2qv(22)d2\begin{aligned} |\vec B|&=\frac{\mu_o}{4 \pi} \frac{q|\vec v|\sin\theta}{|\vec r|^2} \\ |\vec B|&=\frac{\mu_o}{4 \pi} \frac{q(2v)\sin(45^o)}{(\frac{\sqrt{2}}{2}d)^2}\\ |\vec B|&=\frac{\mu_o}{4 \pi} \frac{2qv(\frac{\sqrt{2}}{2})}{(\frac{\sqrt{2}}{2})^2(d)^2}\\ |\vec B|&=\frac{\mu_o}{4 \pi} \frac{2qv}{(\frac{\sqrt{2}}{2})d^2}\\ \end{aligned}
Using the right-hand rule for this charge (point right fingers up the page and curl to the left), we find that the magnetic field is coming out of the page.

Because these two fields have opposite directions, they will partially cancel out. Let's call "out of the page" positive and find the total magnetic field:
∣B⃗net∣=μo4π2qv(22)d2−μo4πqv(22)d2∣B⃗net∣=μo4πqv(22)d2∣B⃗net∣=μo4π2qvd2\begin{aligned} |\vec B_{net}|&=\frac{\mu_o}{4 \pi} \frac{2qv}{(\frac{\sqrt{2}}{2})d^2}-\frac{\mu_o}{4 \pi} \frac{qv}{(\frac{\sqrt{2}}{2})d^2} \\ |\vec B_{net}|&=\frac{\mu_o}{4 \pi} \frac{qv}{(\frac{\sqrt{2}}{2})d^2} \\ |\vec B_{net}|&=\frac{\mu_o}{4 \pi} \frac{\sqrt 2qv}{d^2} \\ \end{aligned}
The result is positive, so the net magnetic field points out of the page.

(Remember that sin⁡45o=22=12\sin45^o=\frac{\sqrt{2}}2=\frac1{\sqrt2}. This means that dividing by 22\frac{\sqrt{2}}2 is the same as multiplying by 2\sqrt2.
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Biot-Savart Law for Currents


Currents are a collection of charges moving together. Each charge contributes to the generation of a magnetic field. Therefore, we can use Biot-Savart to find the total magnetic field generated by a wire.
  • Consider a wire element carrying a current of I and flowing with direction dl⃗d \vec l. The infinitesimal magnetic field contribution from this wire element is given as follows:
dB⃗=μ0I4π dl⃗×r^∣r⃗∣2\boxed{d\vec B= \frac{\mu_0I}{4\pi}\ \frac{{d\vec l}\times \hat{r}} {\left|\vec{r}\right|^2} }
  • The total magnetic field at a point can be found by integrating over all wire elements:
B⃗=∫wiredB⃗\boxed{\vec B=\int_{wire}d\vec B}

Wize Tip
You may also encounter the formula dB⃗=μ0I4π dl⃗×r⃗∣r⃗∣3d\vec B= \frac{\mu_0I}{4\pi}\ \frac{{d\vec l}\times \vec{r}} {\left|\vec{r}\right|^3}. This is identical to the formula given above, only with the position vector replaced by a unit vector.


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Practice: Magnetic Field of a Current Loop


Consider a circular current loop of radius R and current I.

a) Show that the magnitude of the magnetic field generated at some distance z from the center of the loop is given by the following equation.
Bz=μ0IR22(z2+R2)32B_z=\frac{\mu_0IR^2}{2\left(z^2+R^2\right)^{\frac{3}{2}}}
b) What is the magnetic field at the center of the loop?