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3 years ago

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Weak magnetic fields can be measured at the surface of the brain. Although the currents causing these fields are quite complicat

ed, we can estimate their size by modeling them as a current loop around the equator of a 16-cm-diameter (the width of a typical head) sphere. What current is needed to produce a 3.0 pT field—the strength measured for one subject—at the pole of this sphere?

1 answer:

STALIN [3.7K]3 years ago

5 0

To develop this problem it is necessary to apply the concepts related to a magnetic field in spheres.

By definition we know that the magnetic field in a sphere can be described as

$B = \frac{\mu_0}{2}\frac{Ia^2}{(z^2+a^2)^{3/2}}$

Where,

a = Radius

z = Distance to the magnetic field

I = Current

$\mu_0 =$ Permeability constant in free space

Our values are given as

$D=2a = 16cm \rightarrow$ diameter of the sphere then,

$a = 0.08m$

Thus z = a

$B = \frac{\mu_0}{2}\frac{Ia^2}{(a^2+a^2)^{3/2}}$

$B = \frac{\mu_0I}{2(2^{3/2})a}$

$B = \frac{\mu_0 I}{2^{5/2}a}$

Re-arrange to find I,

$I = \frac{2^{5/2}Ba}{\mu_0}$

$I = \frac{2^{5/2}(3*10^{-12})(8*10^{-2})}{4\pi*10^{-7}}$

$I = 1.08*10^{-6}A$

Therefore the current at the pole of this sphere is $1.08*10^{-6}A$

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The question is missing its alternatives. Here is the complete question.

Consider two identical objects released from rest high above the surface of Earth. In Case 1, the object is released from a height above the surface of Earth equal to 1 Earth radius, and we measure its kinetic energy just before it hits the Earth to be K1. In Case 2, the obejct is released from a height above the surface of Earth equal to 2 Earth radii and its kinetic energy just before it hits is K2.

1. Compare the kinetic energy of the two objects just before they hit the surface of the earth.

a) K2 = 2K1; b) K2 = 4K; c) K2 = (4/3)K1; d) K2 = (3/2)K1;

Answer: C) K2 = (4/3)K1

Explanation: As it is related to the gravity of the Earth, the potencial energy is: U(r)= - $\frac{G.Me.m}{r}$ + U₀

In this case, U₀=0, G is the universal gravitational constant, Me is the mass of Earth, m is the mass of the object and r is the distance between the center of the Earth and the object.

The potencial energy of an object of mass m on the surface of the Earth is:

Usurface = - $\frac{G.Me.m}{Re}$

The potencial energy of the object in Case 1 is

U1 = - $\frac{G.Me.m}{2Re}$

For the Case 2:

U2 = - $\frac{G.Me.m}{3Re}$

The potencial change in Case 1:

ΔU1 = - G.Me.m.( $\frac{1}{Re}-\frac{1}{2Re}$ ) = - $\frac{1}{2}$ $\frac{G.Me.m}{Re}$

For Case 2:

ΔU2 = - G.Me.m( $\frac{1}{Re}-\frac{1}{3Re}$ ) = - $\frac{2}{3}$ $\frac{G.Me.m}{Re}$

Comparing ΔK1 and ΔK2 equals comparing ΔU1 and ΔU2:

Δ $\frac{U2}{U1}$ = (-2/3)(-1/2) = 4/3

So, comparing kinetic energies, K2 is 4/3 of K1.

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Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on

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Answer:

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

Explanation:

According to Coulomb's law, the magnitude of force between two point object having change $q_1$ and $q_2$ and by a dicstanced is

$F_c=\frac{1}{4\pi\spsilon_0}\frac{q_1q_2}{d^2}-\;\cdots(i)$

Where, $\epsilon_0$ is the permitivity of free space and

$\frac{1}{4\pi\spsilon_0}=9\times10^9$ in SI unit.

Before dcollision:

Charges on both the sphere are $q_1$ and $q_2$ , d=20cm=0.2m, and $F_c=1.35\times10^{-4}$ N

So, from equation (i)

$1.35\times10^{-4}=9\times10^9\frac{q_1q_2}{(0.2)^2}$

$\Rightarrow q_1q_2=6\times10^{-16}\;\cdots(ii)$

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

$\frac{q_1+q_2}{2}$

d=20cm=0.2m, and $F_c=1.406\times10^{-4}$ N

So, from equation (i)

$1.406\times10^{-4}=9\times10^9\frac{\left(\frac{q_1+q_2}{2}\right)^2}{(0.2)^2}$

$\Rightarrow (q_1+q_2)^2=2.50\times10^{-15}$

$\Rightarrow q_1+q_2=\pm5\times 10^{-8}$

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

$\Rightarrow q_1+q_2=5\times 10^{-8}\;\cdots(iii)$

$\Rightarrow q_1=5\times 10^{-8}-q_2$

The equation (ii) become:

$(5\times 10^{-8}-q_2)q_2=6\times10^{-16}$

$\Rightarrow -(q_2)^2+5\times 10^{-8}q_2-6\times10^{-16}=0$

$\Rightarrow q_2=3\times10^{-8}, 2\times10^{-8}$

From equation (iii)

$q_1=2\times10^{-8}, 3\times10^{-8}$

So, the magnitude of initial charges on both the sphere are $3\times10^{-8}$ Coulombs $=0.03 \mu C$ and $2\times10^{-8}$ Colombs or $0.02 \mu C$ .

Considerion the nature of charges too,

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

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Answer:

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Explanation:

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