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zheka24 [161]
3 years ago
10

A Gaussian surface in the shape of a right circular cylinder with end caps has a radius of 12.5 cm and a length of 112 cm. Throu

gh one end there is an inward magnetic flux of 36.4 μWb. At the other end there is a uniform magnetic field of 2.33 mT, normal to the surface and directed outward. What are the (a) magnitude and (b) direction (inward or outward) of the net magnetic flux through the curved surface?
Physics
1 answer:
snow_tiger [21]3 years ago
8 0

Answer:

A) Magnitude of flux; Φ_c = 77.97 μWb

B) Direction is inwards

Explanation:

A) The magnetic flux through a closed surface is zero. Thus, the summation of the flux through the two bases and the two bases and the curved surface equals zero and is written as:

Φ_s1 + Φ_s2 + Φ_c = 0

Where;

Φ_s1 is the flux through one end and it's given as; -36.4 μWb

(negative sign denotes the direction which is inwards)

Φ_s2 is the flux through the other end is given by a formula;

Φ_s2 = BA

Area is; A = πr²

Thus; Φ_s2 = Bπr²

We are given;

B = 2.33 mT = 2.33 x 10^(-3) T

r = 12.5 cm = 0.125 m

Thus,

Φ_s2 = 2.33 x 10^(-3) x π(0.125²)

Φ_s2 = 114.37 x 10^(-6) Wb

Thus;

Φ_s2 = 114.37 μWb

Since we know Φ_s1 and Φ_s2, let's plug them into the first equation to find the magnetic flux;Φ_c

Thus;

-36.4 μWb + 114.37 μWb + Φ_c = 0

Φ_c = -77.97 μWb

The magnitude is the absolute value which is;Φ_c = 77.97 μWb

B) Since the flux has a negative value, thus the direction is inwards

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Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

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Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

h=\sqrt{(36m)^2+(227m)^2}=229.83m    (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

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yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

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