{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

fall_03_exam2_sol_phy2054

# fall_03_exam2_sol_phy2054 - Solution to Exam 2 Paul Avery...

This preview shows pages 1–3. Sign up to view the full content.

Solution to Exam 2 Paul Avery, Art Hebard PHY2049, Fall 2003 Oct. 20, 2003 1. Two 12 V batteries, A and B, are connected in series. Battery A has an internal resistance of 1 . 0Ω and battery B has an internal resistance of 0 . 5Ω. What is the effective voltage and internal resistance of the combination? Solution: When batteries are put in series, the equivalent EMF is the sum of the voltages and the total resistance is the sum of the individual resistances. Thus the effective EMF is 24V and the internal resistance is 1.5Ω. 2. An infinitely long insulated wire carrying a current I is bent into the shape shown (straight line plus circle of radius R with the currents in the direction shown). The magnitude of the field B at the center of the circle is: I I I R Solution: The total field is the sum of the field from the infinite wire and the circle. Taking into account the fact that the fields are in the same direction, we obtain B = µ 0 I/ 2 πR + µ 0 I/ 2 R = µ 0 I ( π + 1) / 2 πR . 3. Two currents, ﬂow at right angles as shown in the figure. What is the direction of the force on wire segment B due to wire segment A? A B 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Solution: Applying the right hand rule for the field from A, we see that it is going into the page at location B. Using the force right hand rule for the current at B, this gives a force acting in the upwards direction.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}