Answer:
2.64 x 10⁻⁶T
Explanation:
The magnitude of the magnetic field produced by a long straight wire carrying current is given by Biot-Savart law as follows: "The magnetic field strength is directly proportional to the current on the wire and inversely proportional to the distance from the wire". This can be written mathematically as;
B = (μ₀ I) / (2π r) ----------------(i)
B is magnetic field
I is current through the wire
r is the distance from the wire
μ₀ is the magnetic constant = 4π x 10⁻⁷Hm⁻¹
From the question;
I = 0.7A
r = 0.053m
Substitute these values into equation (i) as follows;
B = (4π x 10⁻⁷ x 0.7) / (2π x 0.053)
B = 2.64 x 10⁻⁶T
Therefore the approximate magnitude of the magnetic field at that location is 2.64 x 10⁻⁶T
So, first the formula of Impulse is
I = force * time
We have force but no time.
Then, find time.
Next find acceleration,
F = mass * acceleration
5 = 3 * a
1.67 m/s^2
Next find time,
Acceleration = change in velocity / time
Change in velocity is velocity final - velocity initial
1.67 = 3 - 9 / time
Time = 3.6 s (round to 2 s.f.)
Lastly,
Impulse = force * time
Impulse = 5 * 3.6
Impulse is 18 Ns
The average power output:
P = V * I * t
V = 220 V, I = 15 A;
t = 2 ms * 200 = 400 ms = 0.4 s
P = 220 V * 15 A * 0.4 s
P = 1320 W ≈ 1.3 kW
Answer:
b. 1.3 kW
Answer:
18.1 × 10⁻⁶ A = 18.1 μA
Explanation:
The current I in the wire is I = ∫∫J(r)rdrdθ
Since J(r) = Br, in the cylindrical wire. With width of 10.0 μm, dr = 10.0 μm. r = 1.20 mm. We have a differential current dI. We integrate first by integrating dθ from θ = 0 to θ = 2π.
So, dI = J(r)rdrdθ
dI/dr = ∫J(r)rdθ = ∫Br²dθ = Br²∫dθ = 2πBr²
Now I = (dI/dr)dr at r = 1.20 mm = 1.20 × 10⁻³ m and dr = 10.0 μm = 0.010 mm = 0.010 × 10⁻³ m
I = (2πBr²)dr = 2π × 2.00 × 10⁵ A/m³ × (1.20 × 10⁻³ m)² × 0.010 × 10⁻³ m = 0.181 × 10⁻⁴ A = 18.1 × 10⁻⁶ A = 18.1 μA
