Answer:
0.0025116weber/m²
Explanation:
Magnetic field density (B) is the ratio of the magnetic flux (¶) through the loop to its cross sectional area (A).
Mathematically;
B = ¶/A
¶ = BA
Given B = 0.23Tesla which is the magnitude of the magnetic field
Dimension of the rectangular loop = 7.8 cm by 14 cm
Area of the rectangular loop perpendicular to the field B = 7.8cm×14cm
= 109.2cm²
Converting this value to m²
Area of the loop = 109.2 × 10^-4
Area of the loop = 0.01092m²
Magneto flux = 0.23×0.01092
Magnetic flux = 0.0025116weber/m²
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
Answer:
Interact with it
Explanation:
The body in a condensed form (often a dot or a box)
Straight arrows pointing in the direction in which forces act on the body are represented.
Moments are portrayed by curved arrows pointing in the direction in which they impact the body.
A coordinate system is a series of coordinates.
If the circuit is open, then there will be no light or sound. No current will flow and no air will move.
