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kipiarov [429]
3 years ago
5

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?
Physics
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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A chair of weight 100 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 4
otez555 [7]

Answer:

The normal force will be "122.8 N".

Explanation:

The given values are:

Weight,

W = 100 N

Force,

F = 40 N

Angle,

θ = 35.0°

As we know,

⇒  N=W+FSin \theta

On substituting the given values, we get

⇒      = 100N+40N \ Sin \theta

⇒      =100N+22.8

⇒      =122.8 \ N

7 0
3 years ago
A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh
dem82 [27]

Answer:

Explanation:

a )

Reaction force of the ground

R = mg

= 160 N

Maximum friction force possible

= μ x R

= μ x 160

= .4 x 160

= 64 N .

b )

 160 N will act at middle point . 740N will act at distance of 3 / 5 m from the wall ,

Taking moment about top point of ladder

160 x 1.5 + 740 x 3/5 + f x 4 = 900 x 3

240 + 444 + 4f = 2700

f = 504 N

c )

Let x be the required distance.  

Taking moment about top point of ladder

160 x 1.5 + 740 x 3 x / 5 + .4 x 900 x 4  = 900 x 3  ( .4 x 900 is the maximum friction possible )

240 + 444 x + 1440  = 2700

x = 2.3  m

so man can go upto 2.3 at which maximum friction acts .

8 0
3 years ago
Water is pumped through a pipe of diameter 15.0 cm, from the Colorado River up to Grand Canyon Village, located on the rim of th
Aleksandr [31]

Answer:

p= 1.50289×10⁷ N/m²

Explanation:

Given

HA = (564 m)................(River Elevation)

HB = (2096 m).............(Village Elevation)

Area = A =(π/4){Diameter}² = (π/4){0.15 m}² = 0.017671 m²

ρ = (1 gram/cm³) = (1000 kg/m³)........(Water Density)

p(pressure)=?

Solution

p=PA - PB

p= ρ*g*HB - ρ*g*HA

p= (ρ*g)*(HB - HA)

p= (1000×9.81 )×{2096  - 564}  

p= 1.50289×10⁷ N/m²

8 0
3 years ago
There is much debate about whether or not global warming exists and whether people and their actions are to blame. High levels o
sergiy2304 [10]
<span>C) Humans and their activities do not affect the natural cycles of the Earth

you can think that </span>Humans and their activities do not affect the natural cycles of the Earth.
4 0
3 years ago
Read 2 more answers
When applying a horizontal force of 30N, an object of mass 6 kg accelerates at 4m/s2. The force of friction on the surface must
Rus_ich [418]
Using Newton's Second Law, F = ma, where F is the net force

So the net force is:

F = (6kg)(4m/s^2) = 24N

Since you are applying a horizontal force of 30N, we can find the force of friction by the difference of the net force and the applied force.

30N-24N = 6N

F_{f} = 6N
4 0
3 years ago
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