Download Chapter 27 Problem 51 † Given B = bt k b = 2.1 T/ms t = 0.40 µs

Chapter 27
Problem 51
†
Given
~ = bt k̂
B
b = 2.1 T /ms
t = 0.40 µs = 4.0 × 10−7 s
x = 5.0 cm = 5.0 × 10−2 m
y=0
z=0
~v = 4.8 × 106 ĵ m/s
Solution
Find the net electromagnetic force on the proton.
First convert the value of b into standard units.
2.1 T 1000 ms
= 2100 T /s
b=
ms
1s
At t = 0.40 µ s the magnetic field is
~ = (2100 k̂ T /s)(4.0 × 10−7 s) = 8.4 × 10−4 k̂ T
B
The magnetic force on the proton is then
~
F~m = q~v × B
F~m = (1.6 × 10−19 C)(4.8 × 106 ĵ m/s) × (8.4 × 10−4 k̂ T ) = 6.5 × 10−16 î N
Because the magnetic field is changing, an electric field is generated.
I
~ r = − dΦB
Ed~
dt
The magnetic flux for a circle of radius, r is
~·B
~ = AB cos θ = πr2 B cos θ = (πr2 )bt cos(0) = πr2 bt
ΦB = A
Therefore, dΦB /dt is
dΦB
d((πr2 bt)
=
= πr2 b
dt
dt
When travelling around a circle centerred on the middle of the solenoid, the Electric field is constant and
the line integral becomes
I
I
~
Ed~r = E dr = E2πr
Setting these equal to each other and solving for the electric field gives
E2πr = −πr2 b
E=−
†
πr2 b
rb
(5.0 × 10−2 m)(2100 T m2 /s)
=− =−
= −52.5 N/C
2πr
2
2
Problem from Essential University Physics, Wolfson
Since the magnetic field is initially upward and increasing, the induced electric field is oriented so it
would cause a clockwise current to flow and thus generate a downward magnetic field (by Lenz’s law).
This opposition to change is why the negative sign shows up in the previous calculation. Therefore, at
the location of the proton (x = 5.0 cm y = 0)the electric field vector is
~ = −52.5 ĵ N/C
E
The electric force is then
~ = (1.6 × 10−19 C)(−52.5 ĵ N/C) = −8.4 × 10−18 ĵ N
F~e = q E
The total force on the proton is
F~ = {−8.4 × 10−18 ĵ + 6.5 × 10−16 î} N