Comprehensive Formula sheet - PHYS 272 1 = k = 9 × 10⁹ 4 περ N. m² C² C² le 1.6 x 10-19 C €8.85 x 10-12 μο T. m² 1 x 10-7 N. m² 4 π C m/s 1. The electric field of a point charge q: 2. The electric force on a charge q: F = qE 3. The electric field at a location on the dipole axis: 4 5. 6. 1 2qsr Eaxis ¯ 4π €。 (r−³)² (r+³)² The electric field at a location perpendicular to the dipole axis: 1 q s E₁ = 4 ΠΕ0 · [r² + (²)²] ³/² 3/2 If r >> s, Eaxis 4 περ 1 2qs r3 1 qs If rs, E₁ 4π € г³ For a location on the perpendicular axis of a uniformly charged rod and at a distance r from the center of the rod, the electric field: 1 9 Erod = If r >> L, Erod 4 περ 2+ r () 1 q 4 π Є。 r² 1 2q/L If Lr, Erod 4 περ I The electric field along the axis of a uniformly charged ring, a distance r from the center of the ring: 1 Ering = qr 4πЄo (R2+2)3/2 1 q If r >> R, Ering 4περ μ 1 q 7. The electric potential due to a point charge q: V = 8. 9. The change in the electric potential for a uniform electric field: AV = -E. A The change in the electric potential for non-uniform electric field: AV = -SE. di 10. The change in electric potential energy: AU = q AV 11. The magnetic field of a moving point charge: B = Ho qvxr 4πT 2 12. The magnetic field due to a current-carrying straight wire: 1 If R >> r, Ering qr 4π €。 R³ Ho Bwire = IL Но 21 If L≫r (long wire) Bwire = 4π r√r² + (L/2)² 4π r 13. The magnetic field of a current-carrying coil (A = πR²): μο 2IAN Ho 2IAN B If r> R B coil 4π (r² + R2)3/2' coil 4 π r3 B center 14. The magnetic field inside a long solenoid: HONI B 15. The magnetic field of a bar magnet: Bmagnet 16. Ohms' Law: V = IR 17. Electric Power: P = IV 18. Magnetic force on a moving charge: =qx 19. Magnetic force on a current carrying wire due to a uniform magnetic field: F = ILXB 20. Electric flux for a uniform electric field: * = Ē.ĥ A 21. Electric flux for any electric field: = √ Ē.ñ dA 22. Gauss's Law: 6 E. ñ dA = Σinside Eo 23. Magnetic flux for a uniform magnetic field: += B.nz. 24. Magnetic flux for any magnetic field: += √ B.ñ dA 25. Ampere's Law: $ B. dỉ = HoΣlinside 26. Faraday's Law: emf = - A mag At 27. The frequency of the wave: f = 1/T 28. The angular frequency: v = 2nf 29. The speed of the wave: v = λf 30. Index of refraction: n = c/v 31. Law of refraction (Snell's Law): n₁sine₁ = n₂sine₂ = Ho 2πIN 4TT R ) A long straight wire suspended in the air carries a conventional current of 5.38 A in the +y direction as shown in Figure (the wire runs along the y axis). At a particular instant a proton at location (-0.0212, 0, 0) m has velocity (0,-2×105,0) m/s. (Consider the charge of the proton: qp = 1.6×10-19 C). y ν x + 1. The direction of the magnetic field due to the wire at the location of the proton: OInto the page Out of the page OUpward ODownward 2. What is the (vector) magnetic field due to the wire at the location of the proton? = о 5.075x10^-5 >[T] 3. What is the (vector) magnetic force on the proton due to the current in the wire? F= < -1.624x10^18 0 0 >[N] 4. Now, consider the proton is moving along -z axis with velocity (0, 0, -2×105) m/s, the magnetic force acting on the proton will be: = 0 0 [N]