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4 years ago

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Magnetic Field of a Current Loop: A wire carrying a current is shaped in the form of a circular loop of radius 4.0 mm. If the ma

gnetic field strength at its center is 1.1 mT with no external magnetic fields present, what is the magnitude of the current flowing through the wire?

1 answer:

mr_godi [17]4 years ago

4 0

Answer:

The magnitude of the current flowing through the wire is 7 A.

Explanation:

Given that,

Radius of the circular loop, r = 4 mm = 0.004 m

Magnetic field at the centre of the loop, B = 1.1 mT

We need to find the magnitude of the current flowing through the wire. The magnetic field at the center of the loop is given by the formula as follows :

$B=\dfrac{\mu_oI}{2r}\\\\I=\dfrac{2Br}{\mu_o}\\\\I=\dfrac{2\times 1.1\times 10^{-3}\times 0.004}{4\pi \times 10^{-7}}\\\\I=7\ A$

So, the magnitude of the current flowing through the wire is 7 A.

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Intensity: Radiation of a single frequency reaches the upper atmosphere of the earth with an intensity of 1350 W/m2. What is the

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Answer:

The right answer is "1010 V/m".

Explanation:

The given values are:

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$I=1350 \ W/m^2$

$c = 3.00\times 108 \ m/s$

$\mu_0= 4\pi\times 10^{-7} \ T.m/A$

Now,

The electric field's maximum value will be:

= $\sqrt{2\times u\times c\times I}$

On substituting the values in the above formula, we get

= $\sqrt{2\times 4\times \pi\times 10^{-7}\times 3\times 10^8\times 1350}$

= $\sqrt{32400\times 3.14\times 10^{-7}\times 10^8}$

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a blanket

Explanation:

soft materials absorb sounds meanwhile hard materials just reflect them causing an echo

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A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 34.5 kg.

ozzi

Answer:

$I_1 = 32.5 kg m^2$

Explanation:

Here as we know that total angular momentum is always conserved for child + round system

so here by angular momentum conservation we have

$I_1 \omega_1 = (I_1 + I_2)\omega_2$

here we have

$I_2 = mR^2$ (inertia of boy)

$I_2 = (34.5)(1.45^2)$

$I_2 = 72.5 kg m^2$

now we have

$I_1(2\pi 55) = (I_1 + 72.5)(2 \pi 17)$

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The electric field near the surface of Earth points downward and has a magnitude of 152 N/C. What is the ratio of the magnitude

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Answer:

$2.7\times 10^{12}$

Explanation:

We are given that

Electric field = $E=152N/C$

We have to find the ratio of magnitude of the upward electric force on an electron to the magnitude of gravitational force on the electron.

We know that

F=qE

Charge on an electron,q= $1.6\times 10^{-19}C$

Using the formula

Upward electric force= $152\times 1.6\times 10^{-19}$ N

Upward electric force= $F_e=2.43\times 10^{-17}$ N

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Using the formula

Gravitational force= $F_g=9.1\times 10^{-31}\times 9.8=8.92\times 10^{-30} N$

Ratio of Fe to the Fg= $\frac{2.43\times 10^{-17}}{8.92\times 10^{-30}}$

$\frac{F_e}{F_g}=2.7\times 10^{12}$

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