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