Download Maxwell Equations - Electricity and Magnetism - Solved Past Paper and more Exams Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! 1 Physics 4B FINAL EXAM Chapters 22 - 34 Fall 1997 Name:_________________________________________ Posting Code _____ _____ _____ Solve the following problems in the space provided. Use the back of the page if needed. Each problem is worth 10 points. You must show your work in a logical fashion starting with the correctly applied physical principles which are on the last page. Your score will be maximized if your work is easy to follow because partial credit will be awarded. If you want your grade posted, put alphanumeric characters in the three spaces at the top right. 1. Write down Maxwell’s Equations. Name each one and explain what it means. Gauss's Law for Electricity r E • d r A ∫ = q εo Charges creates diverging electric fields. Faraday's Law r E • d r s ∫ = − dΦB dt Changing magnetic flux creates circulating electric fields. Gauss's Law for Magnetism r B • d r A ∫ = 0 Their are no magnetic monopoles and magnetic fields never diverge. Ampere's Law r B • d r s = µo∫ dq dt +µ oεo dΦe dt Current or changing electric fields create circulating magnetic fields. 2. The charge q1=+12.0µC has a mass of 3.00g and the charge q2=-8.0µC has a mass of 5.00g. At some instant in time they are a distance 3.20cm apart as shown at the right. Find (a)the electric field felt by the charge q1, (b)the electric force on the charge q1, (c)the electric force on the charge q2 and (d)sketch the directions of these vectors in the drawing at the right. (a)The field felt by q1 is created by the point charge q2, E2 = k q2 2r = (9.00x109 ) 8.00x10−6 (0.0320)2 ⇒ E2 = 7.03x10 7N/C (b)The force on q1 can be found from the definition of electric field, F12 = q1E2 = (12.0x10 −6)(7.03x107) ⇒ F12 = 844N (c) The force on q2 is equal and opposite to the force on q1 according to Newton’s Third Law, F21 = 844N q1 q2 3.20cmE2 F12 F 21 2 3. Find the potential due to a ring of total charge Q and radius R at the point P a distance x from the center of the ring along the axis as shown at the right. The potential due to the point charge dq is dV = k dq r ⇒ V = k dq r∫ . For all the dq’s the r is constant so V = k r dq∫ . The sum of all the dq’s is just Q. Therefor V = k Q r . Writing r in terms of R and x gives the answer V = k Q R2 + x2 4. For the circuit shown the current through R3 is measured to be 2.00mA. Find (a)the potential difference across each resistor, (b)the current through each resistor, (c)the current supplied by the battery and (d)the battery voltage. R1=3.00kΩ, R2=9.00kΩ and R3=1.80kΩ. (a)&(b)The voltage across R3 can be found from Ohm’s Rule, V3 = i3R3 = (2.00)(18.0) = 36.0V . Since R2 is in parallel with R3 it must have the same voltage. The current through R2 can now be found from Ohm’s Rule, i2 = V2 R2 = 36.0 9.00 = 4.00mA . Using the junction theorem, the current through R1 must equal the sum of the currents through R2 and R3, i1 = i2 + i3 = 4.00 + 2.00 = 6.00mA . The voltage can be found from Ohm’s Rule, V1 = i1R1 = (6.00)(3.00) = 18.0V . (c)Since R1 is in series with the battery, the current through both must be the same. (d)Using the loop theorem, the battery voltage must equal the sum of the voltages on R1 and R2, V = V1 + V2 = 18.0 + 36.0 = 54.0V . Q R x xP r dq 2R 3R V 1R V(V) I(mA) R(kΩ) 18.0 6.00 3.00 36.0 4.00 9.00 36.0 2.00 18.0 54.0 6.00 battery
