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Bess [88]
3 years ago
8

A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In

itially the field is zero and then changes to 1.50 T, pointing upward when viewed from above, perpendicular to the circular plane, in a time of 130 ms. What is the average induced emf around the border of the circular region?
Physics
1 answer:
Daniel [21]3 years ago
3 0

Answer:81.57\mu V

Explanation:

Given

radius of circular region r=1.50 mm

A=\pi r^2=7.069\times 10^{-6}\ m^2

Magnetic Field B=1.50\ T

time t=130 ms

Flux is given by

\phi =B\cdot A

change in Flux d\phi =(B_f-B_i)A

Emf induced is e=\frac{\mathrm{d} \phi}{\mathrm{d} t}

e=\frac{(1.5)\cdot 7.069\times 10^{-6}}{130\times 10^{-3}}

e=81.57 \mu V

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A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie
Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, V_{Hall} = 20\ mV = 0.02\ V

Magnetic Field, B = 0.10 T

Hall emf, V'_{Hall} = 69\ mV = 0.069\ V

Now,

Drift velocity, v_{d} = \frac{V_{Hall}}{B}

v_{d} = \frac{0.02}{0.10} = 0.2\ m/s

Now, the expression for the electric field is given by:

E_{Hall} = Bv_{d}sin\theta                            (1)

And

E_{Hall} = V_{Hall}d

Thus eqn (1) becomes

V_{Hall}d = dBv_{d}sin\theta

where

d = distance

B = \frac{V_{Hall}}{v_{d}sin\theta}                      (2)

(a) When \theta = 90^{\circ}

B = \frac{0.069}{0.2\times sin90} = 0.345\ T

(b) When \theta = 60^{\circ}

B = \frac{0.069}{0.2\times sin60} = 0.398\ T

5 0
2 years ago
The driver must stop and remain stopped to let a pedestrian cross at a crosswalk when the pedestrian is _____________.
RSB [31]

Answer: the correct option is B ( on the half of the road that the vehicle is traveling).

Explanation: according to Georgia Code About Pedestrians; The driver must stop and remain stopped to let a pedestrian cross at a crosswalk when the pedestrian is

on the half of the road that the vehicle is traveling.

4 0
3 years ago
Read 2 more answers
Mike rides his horse with a constant speed of 20 km/h. How far can he travel in 4 hours?
gizmo_the_mogwai [7]

Answer:

Mike can travel 80 Km in 4 hours

4 0
2 years ago
A 25 kg ball is thrown into the air when thrown it's going 10 m/s calculate how high it travels
Diano4ka-milaya [45]
On Earth, the acceleration of gravity is 9.8 m/s² downward.
So any object with only gravity acting on it gains 9.8 m/s of
downward speed every second.

If the rock starts out moving upward at 10 m/s, then it will
continue upward for only  (10/9.8) = 1.02 second, before
it stops rising and starts falling.

Its average speed during that time is  (1/2) (10 + 0) = 5 m/s .

At an average speed of 5 m/s for 1.02 sec,
the rock rises

                   (5 m/s) x (1.02 sec)  =  5.102 meters .
4 0
3 years ago
a. What proportion of resistors have resistances less than 90 Ω? b. Find the mean resistance. c. Find the standard deviation of
Anettt [7]

Answer:

a) 0.0625 = 6.25%

b) 106.67 Ω

c) 9.43 Ω

d) 1

Explanation:

The probability distribution is given as

f(x) = (x - 80)/800 for 80 < x < 120

f(x) = 0 otherwise.

f(x) = (x/800) - (0.1)

a) Proportion of resistors with resistance less than 90 Ω

P(X < 90) = ∫⁹⁰₈₀ f(x) dx

∫⁹⁰₈₀ f(x) dx = ∫⁹⁰₈₀ [(x/800) - (0.1)]

= [(x²/1600) - 0.1x]⁹⁰₈₀

= [(90²/1600) - 0.1(90)] - [(80²/1600) - 0.1(80)]

= (5.0625 - 9) - [4 - 8]

= -3.9375 + 4 = 0.0625 = 6.25%

b) The mean is given by the expected value expression E(X) = = Σ xᵢpᵢ (with the sum done all over the data set for each variable and its corresponding probability)

It can be written in integral form as

Mean = ∫¹²⁰₈₀ xf(x) dx (with the integral done all over the probability function, i.e. from, 80 to 120)

Mean = ∫¹²⁰₈₀ x[(x/800) - (0.1)] dx

= ∫¹²⁰₈₀ [(x²/800) - (0.1x)] dx

= [(x³/2400) - (0.05x²)]¹²⁰₈₀

= [(120³/2400) - (0.05(120²)] - [(80³/2400) - (0.05(80²)]

= [720 - 720] - [213.33 - 320] = 106.67 Ω

c) Standard deviation = √(variance)

Variance = Var(X) = Σx²p − μ²

μ = mean = expected value = 106.67 Ω

Σx²p = ∫¹²⁰₈₀ x²f(x) dx = ∫¹²⁰₈₀ x² [(x/800) - (0.1)] dx = ∫¹²⁰₈₀ [(x³/800) - (0.1x²)] dx

= [(x⁴/3200) - (0.0333x³)]¹²⁰₈₀

= [(120⁴/3200) - (0.0333(120³)] - [(80⁴/3200) - (0.0333(80)³)]

= (64800 - 57600) - (12800 - 17066.667)

= 11466.667

Variance = 11466.667 - 106.67² = 88.85

Standard deviation = √88.85 = 9.43 Ω

d) Cdf = sum of probabilities over the entire probability function

Cdf = ∫¹²⁰₈₀ f(x) dx = ∫¹²⁰₈₀ [(x/800) - (0.1)] dx

= [(x²/1600) - 0.1x]¹²⁰₈₀ = [(120²/1600) - 0.1(120)] - [(80²/1600) - 0.1(80)] = (9 - 12) - (4 - 8) = -3+4 = 1 as it should be!!!

Hope this Helps!!!

4 0
3 years ago
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