We are on an asteroid with thrusters pointed in 4 directions; Go Left, GO Right, Go Up, and Go Down. We start at coordinate (3,3
). We simultaneously fire all 4 thrusters. Go Left 3N, Go Right 4N, Go Up 2N, and Go Down 5N.
1 answer:
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Hi there!
We can use Biot-Savart's Law for a moving particle:
B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)
q = charge of particle (1.6 × 10⁻¹⁹ C)
μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)
r = distance from particle (2.10 μm)
There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:
Where 'θ' is the angle between the velocity and radius vectors.
a)
To find the angle between the velocity and radius vector, we find the complementary angle:
θ = 90° - 60° = 30°
Plugging 'θ' into the equation along with our other values:
b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.
c)
In this instance, the radius vector and the velocity vector are perpendicular so
'θ' = 90°.
d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.
Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.
Answer:
a = 2.72 [m/s2]
Explanation:
To solve this problem we must use the following kinematics equation:
where:
Vf = final velocity = 1200 [km/h]
Vo = initial velocity = 25 [km/h]
t = time = 2 [min] = 2/60 = 0.0333 [h]
1200 = 25 + (a*0.0333)
a = 35250.35 [km/h2]
if we convert these units to units of meters per second squared
Answer:
(a) 0.345 T
(b) 0.389 T
Solution:
As per the question:
Hall emf,
Magnetic Field, B = 0.10 T
Hall emf,
Now,
Drift velocity,
Now, the expression for the electric field is given by:
(1)
And
Thus eqn (1) becomes
where
d = distance
(2)
(a) When
(b) When