P116B Exam 2 key 101107

# P116B Exam 2 key 101107 - P HYSICS 116B - M ID-T ERM E XAM...

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Unformatted text preview: P HYSICS 116B - M ID-T ERM E XAM 2 A NSWERS - 11 O CTOBER 2007 I. S HORT- ANSWER QUESTIONS ( M ENTAL S ECURITY C ARDS !) (5 POINTS EACH ) 1. ( 5 points ) A battery with an internal resistance of 0.5 generates a current of 2 A by the conversion of chemical to electrical energy. How much power can this battery deliver to an external circuit? [ Be sure to show your work !] The power that the battery can deliver to an external circuit is P = IV = I 2 R , where the resis- tance in this case is the internal resistance of the battery. Since the internal resistance of the battery is 0.5 , the power delivered is 2 W. 2. ( 5 points ) An engineer needs to combine a resistor and a capacitor in such a way that the ca- pacitor will be fully charged in less than 1 s. If the capacitance has a value of 10 F, what resistance should she choose so that the charge on the capacitor reaches 99% of its value in that time? [ Hint : Take e =2.718.] The charge on the capacitor changes at a rate governed by the time constant of the circuit, with a charge given by q t ( ) = E R ( ) 1 " e " t RC ( ) where E is the emf supplied and R is the resistace. (See problem 7 on the exam if you need to refresh your memory.) We want the capacitor to be almost fully charged in less than 1 s, which means that In this case, the term 1 " e " t RC ( ) should equal 0.99. Carrying out the arithmetic, we get 1 " e " t RC ( ) = 0.99 # " e " t RC = " 0.01 # " t RC = l n 0.01 ( ) = " 4.61 # 10 " 6 4.6110 " 5 = R = 0.022 \$ 3. ( 5 points ) A velocity filter (see sketch) is designed so that only the positive ions of 235 U (charge +1.610-19 C, mass 3.910-25 kg) can pass through the ap- paratus without being deflected. If the magnetic field is 1 T pointing into the page as shown in the diagram, what must the value of the electric field be for an ion with velocity v = 10 5 m s to pass straight through? (Give your answer in V/m or N/C.) In order for the ion to pass through undeflected, the force on the ion due to the electric field must just balance the force due to the magnetic field: r F = q r E + q r v " r B = r 0 # r E = \$ r v " r B # r E = 10 5 V m Physics 116B Mid-Term Examination 2 October 11, 2007 Page 2 of 7 Pages 4. ( 5 points ) Using the Biot-Savart law, one can show that the field due to a current flowing along a long straight wire of length 2 a at the P points into the x-y plane, and is given by B y = I 4 " a x 1 x 2 + a 2 ( ) 1 2 Use the binomial approximation (on the list of equations) to show that if a >> x , the field B y is proportional to 1 x . The binomial approximation states that for a small quantity " , 1 + " ( ) n # 1 + n " . We thus need to cast the denominator in the expression for B y in this form. We do this by factoring out the larger of the two quantities a 2 and x 2 to obtain x 2 + a 2 ( ) " 1 2 = 1 x 2 + a 2 ( ) 1 2 = 1 a 1 + x 2 a 2 # \$ % & ( 1 2 = 1 a 1 + x 2 a 2 # \$ % & ( " 1 2 = 1 a 1 " 1 2 x 2 a 2 # \$ % &...
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## This note was uploaded on 04/02/2008 for the course PHYS 116b taught by Professor Haglund during the Fall '07 term at Vanderbilt.

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P116B Exam 2 key 101107 - P HYSICS 116B - M ID-T ERM E XAM...

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