# Magnetic Flux Formula with Solved Problems

In this tutorial, we show you how to use the magnetic flux formula to find flux through loops, solenoids, coils, and other geometries.

## Definition of magnetic flux:

The amount of magnetic field lines passing through a loop of area A which is tilted at angle $\theta$ from the field is defined as magnetic flux and is found by the below formula $\Phi_m=BA \cos \theta$ This definition is similar to that of electric field flux The SI unit of magnetic flux is the weber which can be seen from the formula above as $1 \,weber={\rm 1\, Wb =1\, T\cdot m^2}$ $\theta$ in the magnetic flux formula is the angle between magnetic field vector $\vec{B}$ and the normal (perpendicular) $\hat{n}$ to the plane of the loop.

The above magnetic flux formula is similar to the vector scalar (dot) product of two vectors as $\vec{A}\cdot\vec{B}=AB\cos\theta$.

Thus, the equation for magnetic flux can be written as a dot product of two vectors magnetic field vector $\vec{B}$ and area vector $\vec{A}$ $\Phi_m=\vec{B}\cdot\vec{A}$ where $\vec{A}$, area vector, is defined as a vector perpendicular to the plane of the loop with magnitude equal to the area of the loop $A$.

The SI unit of the area vector is $m^2$.

One of the most important applications of magnetic flux is in Faraday's law of induction

That law states a changing magnetic flux with time creates a voltage.

In the below, using magnetic flux formula some magnetic flux problems are solved for more understanding.

## Magnetic flux formula Problems

Problem (1): A square loop of side 3 cm is positioned in a uniform magnetic field of magnitude 0.5 T so that the plane of the loop makes an angle of $60^\circ$ with the magnetic field. Find the flux passing through the square loop? Solution: Putting the known values into the magnetic flux equation $\Phi_m=BA\cos \theta$, we get \begin{align*}\Phi_m&=BA\cos \theta\\&=(0.5)(0.03\times 0.03)\cos 30^\circ\\&=0.39\quad {\rm mWb}\end{align*} Note that the $\theta$ in the formula above is the angle between $\vec{B}$ and a unit vector perpendicular to the surface.

In this example, the given angle $60^\circ$ is with the surface of the loop not with a vector perpendicular to the surface that is $\hat{n}$.

Problem (2): A circular loop of area $200\,{\rm cm^2}$ sits in the $xz$-plane. Then, a uniform magnetic field of $\vec{B}=0.3\,\hat{i}+0.4\,\hat{j}\, {\rm T}$ applied on it. Determine

(a) Magnitude of the magnetic field
(b) Magnetic flux through the square loop?

Solution
(a) The magnitude of a vector such as $\vec{R}=R_x \hat{i}+R_y\hat{j}$ is found by formula $R=\sqrt{R_x^2+R_y^2}$ so strength (magnitude) of the magnetic field is determined as $B=\sqrt{(0.3)^2+(0.4)^2}=0.5\quad {\rm T}$

(b) This circular loop is positioned at right angle with the $y$ axis so a unit vector perpendicular to it is written as $\hat{n}=\hat{j}$. Now we use the scalar definition of magnetic flux as $\Phi_m=\vec{B}\cdot\hat{n}A$ to find it as below \begin{align*}\Phi_m&=\vec{B}\cdot\hat{n}A\\&=(0.3\hat{i}+0.4\hat{j})\cdot (\hat{j})\left(200\times 10^{-4}\right)\\&=(0.3\,\hat{i}\cdot\hat{j}+0.4\,\hat{j}\cdot\hat{j})(0.2)\\&=0.4\times 0.2\\&=0.08\quad {\rm T}\end{align*} Where in above we used the definition of scalar product of Cartesian unit vectors. In addition, SI unit of area, squared meter, is determined by following conversion rule $\rm{1\, cm^2=10^{-4}\,m^2}$

Problem (3): A rectangular loop of dimensions 3 cm by 5 cm is placed perpendicular in a uniform magnetic field of magnitude 0.1 T. Find the magnetic flux through the loop? Solution: magnetic flux is defined as the product of magnetic field, surface area, and the angle between B and a unit vector perpendicular to the surface with formula $\Phi_m=BA\cos \theta$.

Here, the loop is positioned perpendicular to the magnetic field B namely the angle between B and a unit vector normal to the surface $\hat{n}$ is zero $\theta=0^\circ$. Putting the values in the magnetic flux formula, we get \begin{align*} \Phi_m&=BA\cos \theta \\&=(0.1)(0.03\times 0.05)\cos 0^\circ\\&=15\times 10^{-5}\quad{\rm Wb}\\ or &=0.15\quad{\rm mWb}\end{align*}

Problem (4): Find the magnetic flux through a 100-turn coil of a radius of 10 cm that carries a current of 3.00 A.

Solution: the magnetic flux through one loop of the coil is $\Phi_1=BA\cos \theta$ where B is the magnetic field produced by the loop.

Recall that the magnitude of the magnetic field of a loop is $B=\frac{\mu_0 i}{2 R}$ which is perpendicularly directed into or out of the plane of the loop. Thus, the angle between $\vec{B}$ and the normal to the plane of the loop is $0^\circ$.

Substituting these values into the magnetic flux equation above, we get the flux through one loop as \begin{align*} \Phi_1&=BA\cos \theta \\ \\&=\frac{\mu_0 i}{2 R}\left(\pi R^2\right)\cos 0^\circ\\ \\&=\frac{4\pi\times 10^{-7}\times 3}{(2)(0.10)}\left(\pi\times(0.1)^2\right)\\ \\&=6\pi^2 \times 10^{-8}\quad {\rm Wb}\end{align*} This is the flux passes through one loop of a coil with $N$ such loops, so the total flux through the coil is \begin{align*}\Phi_m&=N\Phi_1\\&=100\times \left(6\pi^2\times 10^{-8}\right)\\&=0.6 \quad {\rm \mu Wb}\end{align*}

Problem (5): Find the magnetic flux through a 1000-turn solenoid of radius 10 cm and a length equal to 30 cm that a current of 3 A passes through it.

Solution: Recall that the magnetic field inside a long solenoid (neglecting the end effects) is uniform with magnitude $B=\mu_0 \frac{N}{L}i$ where N and L are the number of turns and the length of the solenoid, respectively. \begin{align*}B&=\mu_0 \frac NL i\\&=(4\pi\times 10^{-7})\left(\frac{1000}{0.3}\right)(3)\\&=12.56\quad{\rm mT}\end{align*}

In the other hand, each loop of the long solenoid is perpendicular to its uniform magnetic field so $\theta =0^\circ$. The area of one loop is also $A=\pi r^2=\pi(0.1)^2=0.0314\quad {\rm m^2}$ Substituting the above given values into the equation for magnetic flux, we have \begin{align*} \Phi_1&=BA\cos \theta\\&=(12.56\times 10^{-3})(0.0314)\cos 0^\circ\\&=0.4\times 10^{-3}\quad {\rm Wb}\end{align*}This is the flux through each loop of the solenoid, the total flux is found by multiplying it by $N$ turns of the solenoid as below \begin{align*}\Phi_m&=N\Phi_1\\&=(1000)(0.4\times 10^{-3})\\&=0.4\quad {\rm Wb}\end{align*}

Problem (6): A circular loop of a radius of 10 cm is placed perpendicular to a uniform magnetic field with an unknown strength. The magnitude of the flux through the loop is $5\times 10^{-3}\,{\rm T.m^2}$.

(a) What is the magnitude of the magnetic field?
(b) Now, rotate the magnetic field B such that it is parallel to the plane of the loop. How many magnetic field lines pass through the loop?

Solution: the product of magnetic field, surface area, and the angle between $\vec{B}$ and the direction perpendicular to the plane of the surface $\hat{n}$ gives the magnetic flux formula, $\Phi_m=BA\cos \theta$.

(a) B is perpendicular to the surface so the angle between them is zero, $\theta =0$. Substituting the values into the equation above and solving for magnetic field B, we get \begin{align*} \Phi_m&=BA\cos \theta\\ 5\times 10^{-3}&=B (\pi\times (0.1)^2)\cos 0\\ \Rightarrow B&=\frac{5\times 10^{-3}}{\pi \times 10^{-2}}\\ \\&= 0.16 \quad {\rm T}\end{align*}

(b) B is parallel to the loop, namely, it makes an angle of $\theta=90^\circ$ with a vector perpendicular to the loop surface. In this case, the flux through the loop is \begin{align*} \Phi_m&=BA\cos \theta\\&=(0.16)(\pi\times(0.1)^2)\cos 90^\circ\\&=0\end{align*}

Problem (7): A square coil of side length $a=2\,{\rm m}$ is placed in the $xy$-plane in a uniform magnetic field of magnitude 0.1 T as shown in the figure. The direction of the magnetic field is toward the $y$ direction. Find the total magnetic flux through the coil?

Solution: magnetic flux is the scalar product of magnetic field and the area vector with formula $\Phi_m=\vec{B}\cdot\vec{A}$. As you can see, all factors involved in this equation are vectors.

In this example, the magnetic field vector and area vectors are written as $\vec{B}=0.1\,\hat{j},{\rm T}$ and $\vec{A}=4\,\hat{i},{\rm m^2}$. Putting these vectors in the formula above and using the definition of scalar product, we get \begin{align*} \Phi_m&=\vec{B}\cdot\vec{A}\\&=(0.1\,\hat{j})\cdot (4\,\hat{i})\\&=0.4\,(\hat{j}\cdot\hat{i})\\&=0.4\times 0\\&=0\end{align*}

Problem (8): A uniform magnetic field of magnitude B is passing through the base of a hemisphere with radius R. How many field lines penetrate the spherical surface of the hemisphere?

Solution: magnetic flux is a measure of how many magnetic field lines passes through a surface which is computed by the formula $\Phi_m=BA\cos \theta$.

The magnetic flux through the base of the hemisphere is \begin{align*}\Phi_m&=B(A)\cos 180^\circ\\&=-B(\pi R^2)\end{align*} To find the flux through the spherical surface we can use directly the definition of magnetic flux in spherical coordinates (which is hard) or use Gauss's law for magnetism as $\oint\vec{B}\cdot \hat{n}\,dA=0$ According to this law, the magnetic flux through ANY closed surface is always ZERO

A hemisphere is composed of two parts, one is a flat base, and the other spherical part. Total flux $\Phi_m$ through it, which is the sum of the fluxes through base $\Phi_{m-b}$ and spherical $\Phi_{m-S}$, must be zero so the flux passing through the curved surface is obtained as below \begin{align*} \Phi_m&=\Phi_{m-b}+\Phi_{m-S}\\ 0&=-B\pi R^2 +\Phi_{m-S} \\\Rightarrow \Phi_{m-S}&=+B\pi R^2 \end{align*}

Author: Ali Nemati

Page Created: 2/7/2021