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3 years ago

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A 1000-turn coil of wire 1.0 cm in diameter is in a magnetic field that increases from 0.10 T to 0.30 T in 10 ms. The axis of th

e coil is parallel to the field. What is the emf of the coil

1 answer:

ddd [48]3 years ago

4 0

emf generated by the coil is 1.57 V

Explanation:

Given details-

Number of turns of wire- 1000 turns

The diameter of the wire coil- 1 cm

Magnetic field (Initial)= 0.10 T

Magnetic Field (Final)=0.30 T

Time=10 ms

The orientation of the axis of the coil= parallel to the field.

We know that EMF of the coil is mathematically represented as –

E=N(ΔФ/Δt)

Where E= emf generated

ΔФ= change inmagnetic flux

Δt= change in time

N= no of turns*area of the coil

Substituting the values of the above variables

=1000*3.14*0.5*10-4

=.0785

E=0.0785(.2/10*10-3)

=1.57 V

Thus, the emf generated is 1.57 V

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Answer:

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Explanation:

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This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

$q=-kA\frac{dT}{dr}\;\cdots(i)$

where,

K=Thermal conductivity of the material.

T= temperature at any radial distance r.

A=Area through which heat transfer is taking place.

Here, $A=2\pi rL\;\cdots(ii)$

Variation of temperature w.r.t the radius of the annalus is

$\frac {T-T_1}{T_2-T_1}=\frac{\ln(r/r_1)}{\ln(r_2/r_1)}$

$\Rightarrow \frac{dT}{dr}=\frac{T_2-T_1}{\ln(r_2/r_1)}\times \frac{1}{r}\;\cdots(iii)$

Putting the values from the equations (ii) and (iii) in the equation (i), we have

$q=\frac{2\pi kL(T_1-T_2)}{\LN(R_2/2_1)}$

$\Rightarrow k= \frac{q\ln(r_2/r_1)}{2\pi L(T_2-T_1)}$

$\Rightarrow k=\frac{(2.4\pi L)\ln(r_2/r_1)}{2\pi L(10)}$ [as $q=2.4\pi L$ , and $T_2-T_1=10 ^{\circ}C$ ]

$\Rightarrow k=0.12\ln(r_2/r_1)$ $\frac {W}{ m^{\circ} C}$

This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.

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The average pressure on the surface of the wall is the pressure at the centeroid of the equilateral triangular block which can be then be calculated by multiplying it with the Plate Area which will provide us with the Resultant force.

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Noting from the Bernoulli equation that

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From the second image attached the distance of the pressure center from the free surface of the water along the surface of the wall is given by:

$Yp = s+\frac{b}{2} +\frac{b^2}{s+\frac{b}{2}+Po/\rho g sin60}= 0+\frac{28.87}{2} +\frac{28.87^2}{0+\frac{28.87}{2}+100000 /1000 *9.81 sin60} = 17.1 m$

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Answer:

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Explanation:

From the question, we were given the following parameters;

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Rate of energy absorbed is âAG;

âAG = 0.7 × 0.5 × 10000

âAG = 3500W.

The energy absorbed is measured in watts and denoted by the symbol W.

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