Vector b is α so the magnitude of the torque bardbl

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vector B is α , so the magnitude of the torque ( bardbl vector τ bardbl ≡ τ ) is τ = μ B sin θ = I π R 2 B sin α . 004(part1of2)10.0points A particle of mass 3 . 984 × 10 26 kg and charge of 4 . 8 × 10 19 C is accelerated from rest in the plane of the page through a potential difference of 484 V between two parallel plates as shown. The particle is injected through a hole in the right-hand plate into a region of space containing a uniform magnetic field of magnitude 0 . 223 T. The particle curves in a semicircular path and strikes a detector. q m Region of Magnetic Field B E hole Which way does the magnetic field point? 1. out of the page correct 2. toward the bottom of the page 3. to the right 4. toward the upper right corner of the page 5. into the page
dinh (tpd335) – Homework 9 – shubeita – (57460) 3 6. toward the top of the page 7. toward the upper left corner of the page 8. to the left 9. toward the lower right corner of the page 10. toward the lower left corner of the page the Lorentz force law vector F = qvectorv × vector B bardbl vector F bardbl = q v B What is the magnitude of the force exerted on the charged particle as it enters the region of the magnetic field vector B ? = radicalBigg 2 (4 . 8 × 10 19 C) (484 V) (3 . 984 × 10 26 kg) = 1 . 07994 × 10 5 m / s . Let : B = 0 . 223 T . Then the force on the particle is given by the Lorentz force law vector F = qvectorv × vector B bardbl vector F bardbl = q v B
dinh (tpd335) – Homework 9 – shubeita – (57460) 4 7. bardbl vector B CD bardbl = μ 0 I 3 1 b 8. bardbl vector B CD bardbl = 0 9. bardbl vector B CD bardbl = μ 0 I 6 1 b
Solution: For current along arc CD , dvectors is perpendicular to ˆ r . Therefore, if we are not concerned with the field’s direction, the Biot- Savart law gives dB = μ 0 I 4 π b 2 ds for current along arc CD . Integrating over the segment yields B CD = μ 0 I 4 π b 2 integraldisplay D C ds = μ 0 I 4 π b 2 ( b α ) = μ 0 I 4 π b α = μ 0 I 4 π b π 3 = μ 0 I 12 b ,