At SFU the magnetic field due to the earth is at 16.2° to the vertical and has a magnitude of 6.34E-5T. An electron moves straight down at 2.98E+5ms−1. Find the ratio of the magnitude of the magnetic force on the electron to its weight, mg.
Simple question; just plug in numbers
|Fm| / mg = qvBsinθ / mg
q = -1.6E-19 C
m = 9.1E-31 kg
g = 9.81 ms−2
= | (-1.6E-19 C)(2.98E+5ms−1)(6.34E-5T)sin(16.2) | / (9.1E-31 kg)(9.81 m·s−2)
= 9.46×1010 (No units)
Charged Cork Ball
A cork ball carrying charge q has a mass of 2.10 g and is set in motion perpendicular to a uniform magnetic field of 0.90 T. What is the magnitude of q if its direction of motion changes by 5.0° in 1.0 s?
Since the ball is set in motion perpendicular to a uniform magnetic field, it will undergo centripetal motion. In other words, centripetal motion is achieved when the centripetal force equals the magnetic force.
Fc = Fm
mv
q = mv / rBsinθ
To find v, we use the fact that the ball's direction of motion changes by 5.0° in 1.0 s.
ω = v/r where ω = Δθ / Δt = (5*π/180) / 1.0 s (Must convert degrees to radians for angular motion)
ω = 0.0873 s−1
Finally:
q = mv / rBsinθ = (2.10 g * 1 kg / 1000 g)(0.0873 s−1) / (0.90 T)
q = 2.04×10-4 C
Proton Orbit
A proton is sailing through the outer region of the Sun at a speed of 0.275 c. It traverses a locally uniform magnetic field of 0.345 T at an angle of 29.8°. What is the radius of its helical orbit? (Hint: v|| and v⊥ can be considered separately.)
Only v⊥ is influenced by B, so we can completely ignore v||. As such, it becomes a simple plug in values question:
The helical movement is defined the same as circular motion for v⊥:
mv
r = mv / qBsinθ
q = 1.6E-19 C
m = 1.67E-27 kg
c = 3.0E+8 m·s−1
We have enough information to solve for r now:
r = mv / qBsinθ = (1.67E-27 kg)(0.275*3.0E+8 ms−1) / (1.6E-19 C)(0.345 T)sin(29.8)
r = 1.24 m
Current in Parallel Wires
An infinitely long wire lies along the z-axis and carries a current of I=2.95 A in the positive z-direction. A second infinitely long wire is parallel to the z-axis and lies along the plane x=+12.0cm. Find the current in the second wire if the net magnetic field at x=+7.50cm is zero.
Long wire indicates this formula:
B = µ0I / 2πr
µ0 = 4π×10−7 N·A−2
Find B of the first wire, using the distance at which B of the second wire is zero:
B = µ0I / 2πr
B = (4π×10−7 N·A−2)(2.95 A) / (2π)(7.50 cm * 1 m / 100 cm) = 7.87E-6 T
Now we can use this B to find I in the second wire (Using the premise that B1 = B2) by using the distance d - r where d is the separation of the two wires and r is the distance at which B2 = 0
I = B*2πr / µ0
I = (7.87E-6 T)(2π*(12.0 cm - 7.50 cm) * 1 m / 100 cm)) / (4π×10−7 N·A−2)
I = 1.77 A
Parallel Wires
A simplified vector question. Since both wires carry the same amount of current in the same direction, by vector addition, the total B will be 2Bcosθ. The vertical component of each vector cancels, and the horizontal component adds (Since one B points diagonally northwest while the other points diagonally southwest).
The angle θ is found by trigonometry:
arcsin((22.3/2) / 27.04) = 24.35
B = µ0I / 2πr
B = (4π×10−7 N·A−2)(3.90 A) / (2π)(27.04 cm * 1 m / 100 cm) = 2.88E-6 T
By vector addition:
Btotal = 2Bcosθ = 2(2.88E-6 T)cos(24.35)
Btotal = 5.26×10-6 T
What is the direction of the resultant magnetic field at point P? Express its direction, θ, in degrees as follows: a vector pointing left corresponds to 0 degrees, a vector pointing up corresponds to 90 degrees, a vector pointing right corresponds to 180 degrees, and so on.
In the first part, it was determined that the vertical components of the vectors canceled; that means the resultant magnetic field travels parallel to the normal; hence the answer is 0 degrees.
Hydrogen Atom
A rather simplistic model of the hydrogen atom has a single electron revolving around a nuclear proton with an orbital radius of 5.20E-9 m at a speed of 4.00E+6 m/s. Determine the magnetic field at the proton due to the electron.
The electron orbits the proton; this is analogous to the magnetic field at the center of a loop:
B = µ0I / 2r
To find I, recall that:
I = q/T ; current is defined as charge over time. q is proton charge: 1.6E-19 C
To find time T, which is really the period due to circular motion:
T = 2π / ω and ω = v / r so
T = 2π / (v / r) = 2π / (4.00E+6 ms−1 / 5.20E-9 m)
T = 8.17E-15 s
Now find I:
I = q/T = (1.6E-19 C)(8.17E-15 s) = 1.31E-33 C
Finally, find B:
B = µ0I / 2r
B = (4π×10−7 N·A−2)(1.31E-33 C) / 2(5.20E-9 m)
B = 2.37×10-3 T
Small Solenoid
A small diameter, 11- cm long, solenoid has 283 turns and is connected in series with a resistor of 137 Ω. Calculate the magnitude of the magnetic field in the middle of the solenoid when a voltage of 55 V is applied to the circuit.
Simply use the solenoid equation:
B = µ0*N*I / L and I = V / R
B = (4π×10−7 N·A−2)(283)(55 V / 137 Ω) / (11 cm * 1 m / 100 cm)
B = 1.30E-3 T
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