精華區beta NTU-Exam 關於我們 聯絡資訊
課程名稱︰普通物理 課程性質︰必修 課程教師︰蔡爾成 開課系所︰電機系 考試時間︰2005/4/21 試題 : 1.The electric potential at poit (x,y,z) is given by V = (2.0V/m^2)x^2 -(3.0V/m^2)y^2 + (4.0V/m^2)z^2 → What is the electric field E at the point (3.0m, 2.0m ,1.0m)? 2.Find the expression for the oscillation frequency of a magnetic dipole → moment υ and rotational inertial I for small amplitudes of oscillation → about its equilibrium position in a uniform magnetic field B. 3.The following figure shows a long cylindrical conductor of radius a containing a long cylindrical hole of radius b. The central axes of the cylinder and hole are parallel and are distance d apart. Current i is uniformly distributed over the tinted area. Show that the magnetic field in the hole is uniform. 4.An initially uncharged capacitor C is fully charged by a device of constant emf E, in series with a resistor R. Show that the final energy stored in the capacitor is half the energy supplied by the emf device. → → → → 5.Evaluate the torque τ = i∮l × (dl × B ) exerted by a uniform magnetic → field B on a circular loop of radius R lying on the x-y plane with current ^ → i and show that it is equal to iπR^2 k × B. 6.Prove that the static electric field is perpendicular to the equipotentential surface and points to the direction of decreasing electric potential. → 7.Prove that the magnetic field B must be tangent to the surface of a conductor → if B = 0 inside the conductor. 8.At a distance r away from a current loop of finite size carrying a steady current i, show that the magnitude of magnetic field B due to the current loop is proportional to 1/(r^3). 9.Charge Q is uniformly distributed in a spherical region of radius a with uniform charge density ρ = Q/(4/3 πa^3) inside the sphere. What is the electrostatic energy? 10.What is the magnitude of magnetic field produced at the center of a equilateral triangular loop of wire carrying current i? Assume that the length of each side of triangle is a. You may need to make use of the indefinite integral ∫dx/(x^2 + 1 )^1.5 = x/(x^2 + 1 )^0.5. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 220.135.124.228