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Juli2301 [7.4K]
3 years ago
8

People have proposed driving motors with the earth's magnetic field. This is possible in principle, but the small field means th

at unrealistically large currents are needed to produce noticeable torques. Suppose a 19-cmcm-diameter loop of wire is oriented for maximum torque in the earth's field.What current would it need to carry in order to experience a very modest 1.0 x 10³ N/m torque?
Physics
1 answer:
Whitepunk [10]3 years ago
6 0

Answer:

Current needed = 704A

Explanation:

Using the fomula; torque(τ) = (I)(A)(B)Sinθ

Where B = uniform magnetic field

I = current and A = Area

Diameter = 19cm = 0.19m so, radius = 0.19/2 = 0.095m

Area(A) = πr^(2) = πr^(2)

= π(0.095)^(2) = 0.0284 m^(2)

Now, B(earth)= 5x10^-5 T

While, we can ignore the angle because it's insignificant since the angle of the wire is oriented for maximum torque in the earth's field.

Now, if we arrange the formula to solve for charge (I):

I = (τ)/(A)(B)

I = (1.0x10^-3) / (0.0284)(5x10^-5)

I = 704A

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defon
The magnetic force acting on the proton is 
F=qvB \sin \theta
where
q is the proton charge
v is its speed
B is the intensity of the magnetic field
\theta is the angle between the direction of v and B; since the proton is moving perpendicular to the magnetic field, \theta=90^{\circ} and \sin \theta=1, so the force becomes
F=qvB

this force provides the centripetal force that keeps the proton in circular motion:
m \frac{v^2}{r} = q v B
where the term on the left is the centripetal force, with
m being the mass of the proton
r the radius of its orbit

Re-arranging the previous equation, we can find the radius of the proton's orbit:
r= \frac{mv}{qB}= \frac{(1.67 \cdot 10^{-27} kg)(6.5 m/s)}{(1.6 \cdot 10^{-19} C)(1.8 T)}=3.77 \cdot 10^{-8}m

And now we can calculate the centripetal acceleration of the proton, which is given by
a_c =  \frac{v^2}{r}= \frac{(6.5 m/s)^2}{3.77\cdot 10^{-8}m}=1.12 \cdot 10^9 m/s^2

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3 years ago
Which of the following is not part of the process known as oxidative phosphorylation?(a) Molecular oxygen serves as a final elec
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(d) ATP molecules are produced in the cytosol as glucose is converted into pyruvate.

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a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

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<h3>The Radius of the System</h3>

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Learn more on coefficient of static friction here;

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