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1 month ago

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In a demonstration, a 4.00 cm2 square coil with 10 000 turns enters a larger square region with a uniform 1.50 T magnetic field

at a speed of 100 m/s. The plane of the coil is perpendicular to the field lines. If the breakdown voltage of air is 4 000 V/cm on that day, the largest gap you can have between the two wires connected to the ends of the coil and still get a spark is:

1 answer:

serg [3.5K]1 month ago

6 0

Explanation:

It is stated that,

Area of square coil, $A=4\ cm^2=0.0004\ m^2$

Length of one side of the square, L = 0.02 m

Number of coils, N = 10000

Consistent magnetic field, B = 1.5 T

Velocity, v = 100 m/s

An electromotive force is produced in the coil calculated as:

$E=NBLv$

$E=10000\times 1.5\times 0.02\times 100$

E = 30000 V

Breakdown voltage of air, $V=4000\ V/cm=400000\ V/m$

Let d be the distance between the ends of the coil wires that can still produce a spark. Therefore,

Electric field, $E'=\dfrac{V}{d}$

$\dfrac{30000}{d}=400000$

d = 0.075 m

Thus, this forms the final answer.

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Answer:

The final size is nearly the same as the initial size because the increase in size $1.055\times 10^{- 7}$ is remarkably small

Solution:

According to the problem:

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Now,

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This proton energy is $\simeq 250 GeV$

Thus, the speed of the proton, v $\simeq c$

The time to cover 1 km = 1000 m of distance is calculated as:

T = $\frac{1000}{v}$

T = $\frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s$

According to the dispersion factor;

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}$

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}$

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2 months ago

If an instalment plan quotes a monthly interest rate of 4%, the effective annual/yearly interest rate would be _____________. 4%

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Answer:

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2 months ago

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in va

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Answer:

10000 V

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Explanation:

E = Electric field = $4\times 10^6\ V/m$

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The potential difference across the plates amounts to 10000 V

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The area of each plate measures 0.00225988700565 m²

Capacitance is determined by

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