Magnetic Flux

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1 Introduction

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1 Introduction

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The definition of magnetic flux is

$${\Phi }_B=\int \vec{\mathbf{B}}\cdot d\mathbf{A}$$

When the magnitude and direction of $\vec{\mathbf{B}}$ are the same at all points on the surface, the integral simplifies to

$${\Phi }_B=\vec{\mathbf{B}}\cdot \mathbf{A}$$

or, equivalently,

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$${\Phi }_B=BA\cos \varphi$$

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$${\Phi }_B=BA\cos \varphi$$

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where $\varphi$ is the angle between the $\vec{\mathbf{B}}$ and $\vec{\mathbf{A}}$ vectors.

Note that previously, electric flux was covered. The same techniques that apply to computing electric flux apply to magnetic flux. See the Electric Flux activity for additional discussion.

2 Computing ${\Phi }_B$

In the following figure, a square loop of area $A$ that can be rotated about the $z$ axis is shown. Assume that the normal direction of the loop is as shown in the diagram. A uniform external magnetic field points in the $\hat{\mathbf{ȷ}}$ direction.

1. At what angles in the range of $\varphi =[0,360^{\circ }]$ is the magnetic flux zero?

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   Answer: The magnetic flux will be zero when the normal vector is perpendicular to ${\vec{\mathbf{B}}}_{\text{ext}}$. This corresponds to $\varphi =90^{\circ }$ and $\varphi =270^{\circ }$

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2. Draw the loop, $\hat{\mathbf{n}}$, and ${\vec{\mathbf{B}}}_{\text{ext}}$ as they would appear when viewed from above (on the positive $z$–axis) when $\varphi =0^{\circ }$, $\varphi =45^{\circ }$, $\varphi =135^{\circ }$, and $\varphi =180^{\circ }$.

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   Answer:

   

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3. When $\varphi =45^{\circ }$, is the magnetic flux positive or negative? What is its value in terms of a fraction of $B_{o}A$?

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   Answer: From the above diagram, the angle between $\hat{\mathbf{n}}$ and ${\vec{\mathbf{B}}}_{\text{ext}}$ is $\varphi$, so the flux will be positive. Its value will be $B_{o}A\cos 45^{\circ }=B_{o}A/\sqrt{2}$.

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4. When $\varphi =135^{\circ }$, is the magnetic flux positive or negative? What is its value in terms of a fraction of $B_{o}A$?

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   Answer: From the above diagram, the angle between $\hat{\mathbf{n}}$ and ${\vec{\mathbf{B}}}_{\text{ext}}$ is $\varphi$, so the flux will be negative. Its value will be $B_{o}A\cos 135^{\circ }=-B_{o}A\cos 45^{\circ }=-B_{o}A/\sqrt{2}$.

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5. Sketch a plot of the magnetic flux, ${\Phi }_B$, as a function of $\varphi$.

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   From the diagram above, the angle of rotation of the plane is the same as the angle between the $\hat{\mathbf{n}}$ and ${\vec{\mathbf{B}}}_{\text{ext}}$, so the general formula is $\Phi =B_{o}A\cos \varphi$, which is sketched below along with dots that correspond to answers to the above problems.

   

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3 Computing $d{\Phi }_B/dt$

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3 Computing $d{\Phi }_B/dt$

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The time rate of change of magnetic flux through a closed loop is a quantity that will be used when Faraday’s law is covered.

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Suppose the loop in the previous problem rotates at a constant rate such that $\varphi =\omega t$, where $\omega$ is a contant.

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Suppose the loop in the previous problem rotates at a constant rate such that $\varphi =\omega t$, where $\omega$ is a constant.

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1. At what angles in the range $\varphi =[0^{\circ },360^{\circ }]$ is $d{\Phi }_B/dt=0$?

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   The answer to part 5. of the previous problem is ${\Phi }_B=B_{o}A\cos \varphi$, which is sketched below. The slope of ${\Phi }_B$ is zero at $\varphi =0,180^{\circ }$, and $360^{\circ }$.

   

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2. For what angle range is $d{\Phi }_B/dt$ increasing?

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   From the plot of ${\Phi }_B$, the slope is positive for $180^{\circ }<\varphi <360^{\circ }$

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3. For what angle range is $d{\Phi }_B/dt$ decreasing?

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   From the plot of ${\Phi }_B$, the slope is negative for $0^{\circ }<\varphi <180^{\circ }$

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   What is the formula for $d{\Phi }_B(t)/dt$ in terms of $B_o$, $A$, $t$, and $\omega$?

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   ${\Phi }_B=B_{o}A\cos \varphi$, so $d{\Phi }_B/dt=-B_{o}A\omega \sin \omega t$

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