AP Physics C E&M- 13.1 Magnetic Flux- Study Notes- New Syllabus
AP Physics C E&M- 13.1 Magnetic Flux – Study Notes
AP Physics C E&M- 13.1 Magnetic Flux – Study Notes – per latest Syllabus.
Key Concepts:
- Magnetic Flux Through an Arbitrary Area
Magnetic Flux Through an Arbitrary Area
Magnetic flux (\(\Phi_B\)) is a measure of the number of magnetic field lines passing through a given surface.
- It depends on the magnetic field strength, the area, and the orientation of the surface relative to the magnetic field.
General Expression:
\( \mathrm{\Phi_B = \displaystyle \int_A \vec{B} \cdot d\vec{A}} \)
- \(\mathrm{\vec{B}}\) = magnetic field vector
- \(\mathrm{d\vec{A}}\) = differential area vector (magnitude = area element, direction = normal to the surface)
- The dot product means flux depends on the angle between \(\vec{B}\) and the area vector.
Simplified Formula (Uniform Field):
\( \mathrm{\Phi_B = B A \cos \theta} \)
- \(\mathrm{B}\) = magnitude of the magnetic field
- \(\mathrm{A}\) = area of the surface
- \(\mathrm{\theta}\) = angle between the magnetic field and the normal to the surface
Key Ideas:
- If the surface is perpendicular to the field (\(\theta = 0^\circ\)), flux is maximum: \( \Phi_B = BA \).
- If the surface is parallel to the field (\(\theta = 90^\circ\)), flux is zero: \( \Phi_B = 0 \).
- Flux can be positive or negative depending on orientation (convention set by right-hand rule).
- Magnetic flux plays a central role in Faraday’s law of induction: a changing flux induces an emf.
Example:
A uniform magnetic field of \(0.20 \, T\) is directed at an angle of \(30^\circ\) with respect to the normal of a circular loop of radius \(0.10 \, m\). Find the magnetic flux through the loop.
▶️ Answer/Explanation
Step 1: Area of loop: \( \mathrm{A = \pi r^2 = \pi (0.10)^2 = 0.0314 \, m^2} \).
Step 2: Flux formula: \( \mathrm{\Phi_B = B A \cos \theta} \).
Step 3: Substitute: \( \mathrm{\Phi_B = (0.20)(0.0314)(\cos 30^\circ)} \).
Step 4: \( \cos 30^\circ = 0.866 \). \( \mathrm{\Phi_B = 0.20 \times 0.0314 \times 0.866 \approx 5.44 \times 10^{-3} \, Wb} \).
Final Answer: The magnetic flux through the loop is \( \mathrm{5.4 \times 10^{-3} \, Wb} \).
Example:
A square loop of side \(a = 0.20 \, m\) lies in the \(xy\)-plane. A non-uniform magnetic field is directed along the \(z\)-axis and varies with position as \( \mathrm{B(x) = 0.50 \, x} \), where \(x\) is in meters and \(B\) is in tesla. Find the total magnetic flux through the square.
▶️ Answer/Explanation
Step 1: General flux expression: \( \mathrm{\Phi_B = \displaystyle \int_A \vec{B} \cdot d\vec{A}} \).
Step 2: Here, \(\vec{B}\) points along \(z\) and \(d\vec{A}\) also points along \(z\). So \( \vec{B} \cdot d\vec{A} = B(x) \, dA \).
Step 3: The area element is \( dA = dx \, dy \), with \(0 \leq x \leq a\), \(0 \leq y \leq a\).
Step 4: Set up the double integral: \[ \mathrm{\Phi_B = \displaystyle \int_0^a \int_0^a B(x) \, dx \, dy = \int_0^a \int_0^a (0.50x) \, dx \, dy } \]
Step 5: Integrate over \(x\): \(\mathrm{\int_0^a 0.50x \, dx = 0.25 a^2}\).
Step 6: Now integrate over \(y\): \(\mathrm{\int_0^a (0.25 a^2) \, dy = 0.25 a^3}\).
Step 7: Substitute \(a = 0.20 \, m\): \(\mathrm{\Phi_B = 0.25 (0.20)^3 = 0.25 (0.008) = 2.0 \times 10^{-3} \, Wb}\).
Final Answer: The magnetic flux through the square is \( \mathrm{2.0 \times 10^{-3} \, Wb} \).