3 - Electromagnetism ii - Magnetic Fields, Forces

University:
High School
Course:
High School Physics
Academic year:

2024
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499

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8
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3: Electromagnetism II: Magnetic Fields, Forces AP FIZZ Motivation • In Gr 10, we described the “strength” of magnetic fields in solenoids as being proportional to the product of “turns times current”. What is the true strength of magnetic fields around wires and how does a magnetic field affect the motion of charged particles? Strength of Magnetic Fields (Ampere’s Law) • Along any closed path through a magnetic field, the sum of the products of the parallel magnetic field component with the length of that segment, is directly proportional to the net electric field passing through the area enclosed by the path. σ 𝐵 ∆𝑙 = μ𝑜𝐼 || (μ𝑜 = 4π × _7 10 𝑇 · 𝑚/𝐴) Strength of Magnetic Fields (Ampere’s Law) • Choose a circular path of radius, r, around a wire for simplicity, so that path length = circumference (2πr). 𝐵= 𝐼 μ𝑜( ) 2π𝑟 • If solenoid of length L, number of turns N and current I, then 𝐵= 𝑁𝐼 μ𝑜( ) 𝐿 EX 2: What is the magnitude of the magnetic field 2.0 cm from a long, straight conductor with a current of 2.5A? B= 𝐼 μ𝑜 ( ) 2π𝑟 B = (4π × _7 10 )(2.5)/[2π _5 B = 2.5 × 10 𝑇 0.020 ] EX 3: What is the magnitude of the magnetic field in the core of a solenoid 5.0cm long with 300 turns and a current of 8.0A? B= 𝑁𝐼 μ𝑜 ( ) 𝐿 B = (4π × _7 10 )(300)(8.0)/(0.050) _2 B = 6.0 × 10 𝑇 Magnetic Forces on Charged Particles • When charges, q, are moving at a velocity, v, in a magnetic field of strength, B, where angle between v and B is θ, force on particle is given by 𝐹𝑀 = 𝑞𝑣𝐵 sin θ (unit of B is Tesla (T), 1T = 1kg/C·s) • If those charges are in a conductor of length 𝑙 inside the magnetic field and carry a current 𝐼, then the force is 𝐹𝑀 = 𝐼𝑙𝐵 sin θ (dir perpendicular to both v and B, or l and B -> circular motion!) EX 1:Electron accelerates from rest in horizontal electric field, V=46V. Then leaves electric field, enters magnetic field 0.20T into page. Find ve, FM, and r of the circular path of electron. 1 𝑞𝑉 = 𝑚𝑣2 2 _ 𝑣= 2(1.6 × 10 19)(46) = 4.0 × 106 𝑚/𝑠 _31 9.11 × 10 _ _ 𝐹𝑀 = 𝑞𝑣𝐵 sin 90 = 1.6 × 10 19 4.0 × 106 0.20 = 1.3 × 10 13𝑁 𝐹𝑀 = 𝐹𝐶 𝑚𝑣 2 𝑞𝑣𝐵 = 𝑟 _31 𝑚𝑣 (9.11 × 10 )(4.0 × 106) _ 𝑟= = = 1.1 × 10 4 𝑚 _19 𝑞𝐵 (1.6 × 10 )(0.20)