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malfutka [58]
3 years ago
5

A11) A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic fiel

d strength is inside the solenoid near its center when a certain current flows through the coils. If the coils of the solenoid are now pulled apart slightly, stretching it to without appreciably changing the size of the coils, what does the magnetic field become near the center of the solenoid when the same current flows through the coils? (μ0 = 4π × 10-7 T • m/A) A) 1.7 mT B) 3.4 mT C) 0.85 mT D) 2.0 mT
Physics
1 answer:
vitfil [10]3 years ago
7 0

Answer:

A

Explanation:

From a Solenoid we know that a magnetic fiel is always inversely proportional to lenght L or BL = constant

B= frac{\mu_0}{2R}

As I is constant

\frac{B2}{B1} = \frac{R1}{R2}

B2 = 2mT*\frac{18}{21}

B2 = 1.714mT

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A gun shoots a bullet with a velocity of 500 m/s. The gun is aimed horizontally and fired from a height of 1.5 m. How far does t
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The bullet travels a horizontal distance of 276.5 m

The bullet is shot forward with a horizontal velocity u_x. It takes a time <em>t</em> to fall a vertical distance <em>y</em> and at the same time travels a horizontal distance <em>x. </em>

The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.

The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.

Calculate the time taken for the bullet to fall through a vertical distance <em>y </em>using the equation,

y=u_yt+\frac{1}{2} gt^2

Substitute 0 m/s for u_y, 9.81 m/s²for <em>g</em> and 1.5 m for <em>y</em>.

y=u_yt+\frac{1}{2} gt^2\\ 1.5 m=(0m/s)t+\frac{1}{2} (9.81m/s^2)t^2\\ t=\sqrt{\frac{2(1.5m)}{9.81m/s^2} } =0.5530s

The horizontal distance traveled by the bullet is given by,

x=u_xt

Substitute 500 m/s for u_x and 0.5530s for t.

x=u_xt\\ =(500m/s)(0.5530s)\\ =276.5m

The bullet travels a distance of 276.5 m.


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