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SpyIntel [72]
3 years ago
15

A 200-loop coil of cross sectional area 8.5 cm2 lies in the plane of the paper. Directed out of the plane of the paper is a magn

etic field of 0.06 T. The field out of the paper decreases to 0.02 T in 12 milliseconds. What is the direction of the current induced?
Physics
1 answer:
Maslowich3 years ago
5 0

Explanation:

It is given that,

Number of turns, N = 200

Area of cross section, A = 8.5 cm²

Magnetic field is directed out of the paper and is, B = 0.06 T

The magnetic field is  out of the paper decreases to 0.02 T in 12 milliseconds. We need to find the direction of current induced. The induced emf is given by :

\epsilon=-N\dfrac{d\phi}{dt}

Since, \epsilon=IR

I is the induced current

I=-\dfrac{N}{R}\dfrac{d\phi}{dt}

According to Lenz's law, the direction of induced current is such that it always opposes the change in current that causes it.

Here, the field is directed out of the plane of the paper, this gives the induced current in counterclockwise direction.

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Which phenomena support only the wave theory of light? select 2 options. reflection refraction diffraction interference photoele
attashe74 [19]

Interference and diffraction are the phenomena that support only the wave theory of light. Options 2 and 3 are correct.

<h3 /><h3>What is the interference of waves?</h3>

The result of two or more wave trains flowing in opposite directions on a crossing or coinciding pathways. This phenomenon is known as the interference of waves.

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2 years ago
The acclaimed environmental book, Silent Spring, was written by this woman in 1962.
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7 0
3 years ago
The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete
oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\

make shear stress (τ) the subject of the formula

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of  0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2

Part (b) shear stress at a distance of  0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0}{2*12} =0

3 0
4 years ago
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