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LAB 5- University of California, San Diego PHYS 2CL

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PHYS 2CL LAB 5 Worksheet Section 5 1. Consider a coil of radius, rcoil, placed on a magnet which has strength, Bcenter, at the center of the coil, and Bouter at the coil radius. Write the equation for a line, B(r), which takes the value, Bcenter, at r = 0 and the value, Bouter, at r = rcoil. B(r)=Bcenter−( Bcenter−Bouter r coil ) r 2. Integrate the function you found in the previous question over the area of the coil to find the flux, using the general polar surface integration formula below. Φ=∫ 0 2π dϕ∫ 0 rcoil B(r) rdr Given that B(r)=Bcenter−( Bcenter−Bcoil r coil ) r , integrating the above equation gives: Φ=∫ 0 2π dϕ∫ 0 rcoil( Bcenter−( Bcenter−Bouter r coil ) r) rdr Φ=∫ 0 2 π dϕ∫ 0 rcoil( Bcenter r−( Bcenter−Bouter rcoil ) r 2 ) dr Φ=∫ 0 2π dϕ(( −Bcenter+Bouter rcoil ) r 3 3 +Bcenter r 2 2 ) ¿0 rcoil ϕ∨¿0 2π (( −Bcenter+Bouter 3rcoil ) rcoil 3 + Bcenter 2 rcoil 2 ) Φ=¿ Φ=2π rcoil 2 [ ( Bouter−Bcenter) 3 + Bcenter 2 ] Lab Analysis Section 5 Measured Resistance

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Alexander Lee
PHYS 2CL

LAB 5
Worksheet

Section 5

1. Consider a coil of radius, rcoil, placed on a magnet which has strength, Bcenter, at the center of
the coil, and Bouter at the coil radius. Write the equation for a line, B(r), which takes the value,
Bcenter, at r = 0 and the value, Bouter, at r = rcoil.

Bcenter −B outer
B ( r )=Bcenter − ( r coil
r )
2. Integrate the function you found in the previous question over the area of the coil to find the
flux, using the general polar surface integration formula below.
2π rcoil

Φ=∫ dϕ ∫ B ( r ) rdr
0 0
B −B coil
Given that (
B ( r )=Bcenter − center
r coil )
r , integrating the above equation gives:
2π rcoil
Bcenter −B outer
Φ=∫ dϕ ∫ B center −
0 0
( ( r coil

r rdr
rcoil
))
B center −B outer 2
Φ=∫ dϕ ∫ B center r−


0 0
( ( r coil )) r dr

−Bcenter + Bouter r 3 r2 r
Φ=∫ dϕ
0
(( r coil 3 )
+ Bcenter
2 0
¿
) coil




−B center + Bouter 3 Bcenter 2
ϕ∨¿ 20 π
(( 3 r coil
r coil +
) r
2 coil )
Φ=¿
( B outer −B center ) Bcenter
Φ=2 π r 2coil [ 3
+
2 ]
Lab Analysis
Section 5

Measured Resistances and Capacitance:




This study source was downloaded by 100000850872992 from CourseHero.com on 05-27-2023 23:04:52 GMT -05:00


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, Capacitance of 1µF capacitor: 1.006 ± 0.001 µF

Resistance of 100kΩ Decade Box Resistor: 101.2 ± 0.1 kΩ



Measured Diameter Bounds and Calculation of Average Radius with Uncertainty:


dinner = 1.90 ± 0.01 cm

douter = 3.85 ± 0.01 cm

1.90 cm+3.85 cm
ravg = 2
=1.4375 cm
2

0.01 cm 2 0.01 cm 2
δr =
avg
√( 1.90 cm )(
+
3.85 )
( 1.44 cm) =0.008451626 …


Thus, ravg = 1.438 ± 0.008 cm

Table of Voltage Measurements:



Integrator Circuit Peak Voltage Measurements

V1 0.355 V V11 0.392 V
V2 0.469 V V12 0.416 V
V3 0.467 V V13 0.459 V
V4 0.484 V V14 0.426 V
V5 0.473 V V15 0.376 V
V6 0.468 V V16 0.383 V
V7 0.417 V V17 0.475 V
V8 0.407 V V18 0.434 V
V9 0.423 V V19 0.407 V
V10 0.523 V V20 0.377 V

Calculation of average voltage and standard deviation.




This study source was downloaded by 100000850872992 from CourseHero.com on 05-27-2023 23:04:52 GMT -05:00


https://www.coursehero.com/file/63108620/LAB-5docx/

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