Motion in an magnetic field IB DP Physics Study Notes - 2025 Syllabus
Motion in an magnetic field IB DP Physics Study Notes
Motion in an magnetic field IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand
The force per unit length between parallel wires as given by \(\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}\) where \( r \) is the separation between the two wires.
Standard level and higher level: 6 hours
Additional higher level: There is no additional higher level content.
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Force on a current-carrying conductor in a B-field
EXAMPLE:
Explain the colors of the Aurora Borealis, or northern lights.
SOLUTION:
∙The aurora borealis is caused by the interaction of charged particles from space with the earth’s magnetic field, and their subsequent collisions with N2 and O2 molecules in the upper atmosphere.
∙Nitrogen glows violet and oxygen glows green during the de-ionization process.
EXAMPLE: Explain the source of the charged particles causing the Aurora Borealis.
EXAMPLE:
Explain why the Aurora Borealis occurs near the north pole.
SOLUTION:
∙Because the charged particles are moving, the earth’s magnetic field causes a force on them that brings them spiraling into the upper atmosphere where they ionize oxygen and nitrogen molecules.
We can use this special case of parallel current carrying wires and Ampere’s Force Law:
Where L is the length in parallel, r is the distance between the wires, μ₀ is the permeability of free space (4π $\times $ 10-7 T m A-1), and I₁ and I₂ are the currents in the wires.
PRACTICE:
Two parallel 10 cm wires are shown. The given currents in Wire 1 is 0.50 A. What is the Force acting on the wires if they are 5.0 cm apart?
▶️Answer/Explanation
SOLUTION:
\(\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi r}\)
Substituting the given values:
\(\frac{F}{L} = \frac{(4\pi \times 10^{-7} \, \text{T·m/A})(0.50 \, \text{A})(1.5 \, \text{A})}{2 \pi (0.050 \, \text{m})}\)
Simplify the terms:
\(\frac{F}{L} = \frac{4\pi \times 10^{-7} \times 0.50 \times 1.5}{2\pi \times 0.050}\)
\(\frac{F}{L} = \frac{3.0 \times 10^{-7}}{0.10}\)
\(\frac{F}{L} = 3.0 \times 10^{-6} \, \text{N/m}\)
Now, considering a length \( L = 1 \, \text{m} \), the force is:
\(F = 3.0 \times 10^{-6} \, \text{N}\)
This small force is why wires don’t tend to fly together or apart.