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4vir4ik [10]
3 years ago
5

A wire with a current of 1.31 A is to be formed into a circular loop of one turn. If the required value of the magnetic field at

the center of the loop is 10 µT, what is the required radius?
Physics
1 answer:
Sliva [168]3 years ago
4 0

Answer:

The required radius is 2.62 cm

Explanation:

Given;

magnitude of current in wire, I = 1.31 A

magnetic field strength, B =  10 µT

Applying Biot Savart equation;

B = \frac{\mu_o I}{2\pi r} \\\\r =  \frac{\mu_o I}{2\pi B}

where;

r is the radius of the wire

μ₀ is constant = 4π x 10⁻⁷ Tm/A

I is the current in the wire

B is the magnetic field strength

Substitute the given values and calculate the radius

r =  \frac{4 \pi *10^{-7} *1.31}{2\pi *10 *10^{-6}} = 2.62 *10^{-2} \ m = 2.62 \ cm

Therefore, the required radius is 2.62 cm

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The angular speed of the rotor in a centrifuge increases from 420 to 1420 rad/s in a time of 5.00 s. (a) Obtain the angle throug
9966 [12]

Answer:

a) θ = 2500 radians

b) α = 200 rad/s²

Explanation:

Using equations of motion,

θ = (w - w₀)t/2

θ = angle turned through = ?

w = final angular velocity = 1420 rad/s

w₀ = initial angular velocity = 420

t = time taken = 5s

θ = (1420 - 420) × 5/2 = 2500 rads

Again,

w = w₀ + αt

α = angular accelaration = ?

1420 = 420 + 5α

α = 1000/5 = 200 rad/s²

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4 years ago
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Nasa’s goal is to help industry reduce emissions from aircraft by how much by 2050 compared to 2005?
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Answer:

50%

Explanation:

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1. Halve the carbon emission due to aircraft by 2050, relative to emission of 2005

2. To make growth of industry carbon neutral by 2020

3. To cut CO₂ emission by 1.5% per year till 2020

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8 0
3 years ago
How much energy is stored in a 2.80-cm-diameter, 14.0-cm-long solenoid that has 200 turns of wire and carries a current of 0.800
ozzi

Answer:

The energy stored in the solenoid is 7.078 x 10⁻⁵ J

Explanation:

Given;

diameter of the solenoid, d = 2.80 cm

radius of the solenoid, r = d/2 = 1.4 cm

length of the solenoid, L = 14 cm = 0.14 m

number of turns, N = 200 turns

current in the solenoid, I = 0.8 A

The cross sectional area of the solenoid is given as;

A = \pi r^2\\\\A = \pi (0.014)^2\\\\A = 6.16*10^{-4} \ m^2

The inductance of the solenoid is given by;

L = \frac{\mu_0 N^2A}{l} \\\\L =  \frac{(4\pi*10^{-7})(200^2)(6.16*10^{-4})}{0.14}\\\\L = 2.212*10^{-4} \ H

The energy stored in the solenoid is given by;

E = ¹/₂LI²

E = ¹/₂(2.212 x 10⁻⁴)(0.8)²

E = 7.078 x 10⁻⁵ J

Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J

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3 years ago
All of the following are true about scientific evidence except: *
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Answer:

Your answer is D It does not need to be repeatable.

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