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Taya2010 [7]
3 years ago
8

Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 21.0 cm and carries a

clockwise current of 16.0 A , as viewed from above, and the outer wire has a diameter of 32.0 cm.
Required:
a. What must be the direction (as viewed from above) of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?
b. What must be the magnitude of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?
Physics
2 answers:
likoan [24]3 years ago
5 0

Answer:

(a) counter clockwise

(b) 24.38 A

Explanation:

inner diameter, d = 21 cm

inner radius, r = 10.5 cm

Current in inner loop, I = 16 A clock wise

Outer diameter, D = 32 cm

Outer radius, R = 16 cm

(a) The magnetic filed due to the inner wire is inwards to the plane of paper. According to the Maxwell's right hand thumb rule, the direction of magnetic field in outer wire should be outwards so that the net magnetic field is zero at the center.

So, the direction of current in outer wire is counter clock wise in direction.

(b) Let the current in outer wire is I'.

The magnetic field due to the inner wire is balanced by the magnetic field due to the outer wire.

\frac{ \mu 0}{4\pi}\times \frac{2 I}{r}=\frac{\mu 0}{4\pi}\times \frac{2 I'}{R}\\\frac{16}{10.5}=\frac{I'}{16}\\\\I' = 24.38 A

Burka [1]3 years ago
3 0

Solution :

a). B at the center :

     $=\frac{u\times I}{2R}$

Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.

Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE

b). Also, the sum of the fields must be zero.

Therefore,

$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$

So,

$\frac{I_1}{d_1}= \frac{I_2}{d_2}$

$=\frac{16}{21}=\frac{I_2}{32}$

$I_2=24.38 $ A

Therefore, the current in the outer wire is 24.38 ampere.

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Answer:

Explanation:

Diameter of pool = 12 m

radius of pool, r = 6 m

Total height raised, h = 3 + 2.5 = 5.5 m

density of water, d = 1000 kg/m³

Mass of water, m = Volume of water x density

m = πr²h x d

m = 3.14 x 6 x 6 x 5.5 x 1000

m = 113040 kg

Work = m x g x h

W = 113040 x 9.8 x 5.5

W = 6092856 J

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3 years ago
Derive the formula for the moment of inertia of a uniform, flat, rectangular plate of dimensions l and w, about an axis through
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Answer:

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in (Figure). We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

λ = m/l (orm) = λl

If we take the differential of each side of this equation, we find

d m = d ( λ l ) = λ ( d l )

since  

λ

is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact,  

d l = d x

in this situation. We can therefore write  

d m = λ ( d x )

, giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain

I=∫r2dm=∫x2dm=∫x2λdx.

The last step is to be careful about our limits of integration. The rod extends from x=−L/2x=−L/2 to x=L/2x=L/2, since the axis is in the middle of the rod at x=0x=0. This gives us

I=L/2∫−L/2x2λdx=λx33|L/2−L/2=λ(13)[(L2)3−(−L2)3]=λ(13)L38(2)=ML(13)L38(2)=112ML2.

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2 years ago
Adhira loves to ride her bike around the neighborhood. She starts riding 1.2 miles at 30° S of E. Then, she rides another 2.0 mi
zepelin [54]

Answer:

D = 1.8677 miles , θ = 24.28º at South of West

Explanation:

This is an exercise in adding vectors, the easiest way to solve them is to decompose the vectors and add each component algebraically. Let's use trigonometry

first displacement. d = 1.2 miles to 30º south of East

     cos ( 360-30) = cos (-30) = x₁ / d

     sin (-30) = y₁ / d

     x₁ = d cos (-30)

     y₁ = d sin (-30)

     x₁ = 1.2 cos (-30) = 1,039 miles

     y₁ = 1.2 sin (-30) = -0.6 miles

second shift. d = 2.0 miles to 20º West of South

       cos (270-20) = x₂ / d

       cos (250) = y₂ / d

       x₂ = 2.0 cos 250 = -0.684 miles

       y₂ = 2.0 sin250 = -1.879 miles

Third displacement. d = 1.6 miles to 30º South of West

       cos (180 + 30) = x₃ / d

       sin (210) = y₃ / d

       x₃ = 1.6 cos 210 = -1.3856 miles

       y₃ = 1.6 sin 210 = -0.8 miles

Fourth displacement. d = 2.6 miles to 15º West of North

       cos (90 + 15) = x₄ / d

       sin (105) = y₄ / d

       x₄ = 2.6 cos 105 = -0.6729 miles

       y₄ = 2.6 sin 105 = 2,511 miles

having all the components we add

x-axis  (West-East direction)

       X = x₁ + x₂ + x₃ + x₄

       X = 1.039 -0.684 - 1.3846 - 0.6729

       X = -1.7025 miles

   

       Y = y₁ + y₂ + y₃ + y₄

       Y = -0.6 -1.879 -0.8 +2.511

       Y = -0.768

The modulus of this displacement is we use the Pythagorean theorem

      D = √ (X² + Y²)

      D = √ (1.7025² + 0.768²)

      D = 1.8677 miles

let's use trigonometry to find the direction

       tan θ = Y / X

       θ = tan⁻¹ Y / x

       θ = tan⁻¹ (0.768 / 1.7025)

       θ = 24.28º

as the two components are negative this angle is in the third quadrant

therefore in cardinal direction form is

         θ = 24.28º at South of West

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