AP Physics C E&M- 13.3 Induced Currents and Magnetic Forces- Study Notes- New Syllabus
AP Physics C E&M- 13.3 Induced Currents and Magnetic Forces – Study Notes
AP Physics C E&M- 13.3 Induced Currents and Magnetic Forces – Study Notes – per latest Syllabus.
Key Concepts:
- Force on a Conductor Due to Induced Current in a Magnetic Field
Force on a Conductor Due to Induced Current in a Magnetic Field
When a conductor moves in a magnetic field, a changing flux induces a current in the conductor (Faraday’s Law). The induced current interacts with the external magnetic field, producing a force on the moving charges inside the conductor. This is a direct consequence of the Lorentz force acting on charge carriers.
When an induced current is created in a conductive loop, the already-present magnetic field will exert a magnetic force on the moving charge carriers within the loop.
Relevant Equation (Magnetic Force on a Current-Carrying Wire):
$\mathrm{\vec{F}_B = \displaystyle \int I ,(d\vec{l} \times \vec{B})}$
\( \mathrm{\vec{F} = I \, \vec{L} \times \vec{B}} \)
- \( \mathrm{I} \) = induced current
- \( \mathrm{\vec{L}} \) = length vector of the wire in the field
- \( \mathrm{\vec{B}} \) = external magnetic field
- Direction: determined by the right-hand rule
Key Ideas:
The induced emf depends on the rate of change of flux:
\( \mathrm{\mathcal{E} = -\dfrac{d\Phi_B}{dt}} \).
Induced current is given by Ohm’s law:
\( \mathrm{I = \dfrac{\mathcal{E}}{R}} \), where \(R\) is the resistance of the loop.
- The resulting magnetic force on the loop is proportional to this induced current and the external field.
- Only the parts of the loop physically inside the magnetic field experience magnetic forces.
The forces may cause:
- Translational motion (linear acceleration of the loop).
- Rotational motion (torque on the loop).
- Newton’s Second Law applies: \( \mathrm{\sum \vec{F} = m \vec{a}} \).
Energy Consideration:
- The magnetic force that opposes motion of the conductor ensures that mechanical work is converted into electrical energy (current and resistive heating in the loop).
- This is consistent with the principle of conservation of energy.
Example:
A rectangular conducting loop of resistance \(0.50 \, \Omega\) moves with constant velocity \(2.0 \, m/s\) into a region with a uniform magnetic field of strength \(0.40 \, T\). The side of the loop perpendicular to the velocity is \(0.10 \, m\). Find the magnitude of the magnetic force acting on the loop while it is entering the field.
▶️ Answer/Explanation
Step 1: Induced emf: \( \mathrm{\mathcal{E} = B L v = (0.40)(0.10)(2.0) = 0.080 \, V} \).
Step 2: Induced current: \( \mathrm{I = \dfrac{\mathcal{E}}{R} = \dfrac{0.080}{0.50} = 0.16 \, A} \).
Step 3: Magnetic force on side of length \(L\): \( \mathrm{F = I L B = (0.16)(0.10)(0.40) = 6.4 \times 10^{-3} \, N} \).
Final Answer: The force on the loop is \( \mathrm{6.4 \times 10^{-3} \, N} \), opposing the motion of the loop (Lenz’s Law).
Example :
A conducting rod of length \(0.25 \, m\) slides to the right at a speed of \(3.0 \, m/s\) on two parallel frictionless conducting rails in a region with a uniform magnetic field of strength \(0.60 \, T\) directed into the page. The rails are connected by a resistor of \(2.0 \, \Omega\). Find the magnitude and direction of the magnetic force on the rod.
▶️ Answer/Explanation
Step 1: Induced emf across the rod: \( \mathrm{\mathcal{E} = B L v = (0.60)(0.25)(3.0) = 0.45 \, V} \).
Step 2: Induced current: \( \mathrm{I = \dfrac{\mathcal{E}}{R} = \dfrac{0.45}{2.0} = 0.225 \, A} \).
Step 3: Magnetic force on the rod: \( \mathrm{F = I L B = (0.225)(0.25)(0.60) = 0.0338 \, N \approx 3.4 \times 10^{-2} \, N} \).
Step 4 (Direction): – Current flows along the rod (from one rail to the other). – Using the right-hand rule (\(\vec{F} = I \vec{L} \times \vec{B}\)), the force is directed to the left, opposing the rod’s motion (consistent with Lenz’s Law).
Final Answer: The magnetic force is \( \mathrm{3.4 \times 10^{-2} \, N} \), directed to the left.