Answer:
a) B = 1.99 x 10⁻⁴ Tesla
b) B = 0.88 x 10⁻⁴ Tesla
Explanation:
According to Biot - Savart Law, the magnetic field due to a currnt carrying straight wire is given as:
B = μ₀ I L/4πr²
where,
μ₀ = permebility of free space = 1.25 x 10⁻⁶ H m⁻¹
I = current = 2 A
L = Length of wire = 40 cm = 0.4 m
a)
r = radius of magnetic field = 2 cm = 0.02 m
Therefore,
B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.02 m)²
<u>B = 1.99 x 10⁻⁴ Tesla</u>
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b)
r = radius of magnetic field = 3 cm = 0.03 m
Therefore,
B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.03 m)²
<u>B = 0.88 x 10⁻⁴ Tesla</u>
Answer:
V_{a} - V_{b} = 89.3
Explanation:
The electric potential is defined by
= - ∫ E .ds
In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.
V_{b} - V_{a} = - ∫ E ds
We substitute
V_{b} - V_{a} = - ∫ (α + β/ y²) dy
We integrate
V_{b} - V_{a} = - α y + β / y
We evaluate between the lower limit A 2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m
V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)
V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33
V_{b} - V_{a} = - 89.3 V
As they ask us the reverse case
V_{b} - V_{a} = - V_{b} - V_{a}
V_{a} - V_{b} = 89.3
The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m
s = ut + 1 / 2 at²
s = Distance
u = Initial velocity
t = Time
a = Acceleration
Vertically,
s = 15.4 m
u = 0
a = 9.8 m / s²
15.4 = 0 + ( 1 / 2 * 9.8 * t² )
t² = 3.14
t = 1.77 s
Horizontally,
u = 3.01 m / s
a = 0 ( Since there is no external force )
s = ( 3.01 * 1.77 ) + 0
s = 5.34 m
Therefore, the distance travelled by the ball before hitting the ground is 5.34 m
To know more about distance travelled
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