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snow_lady [41]
3 years ago
10

Consider a solenoid of length L, N windings, and radius b (L is much longer than b A current I). is flowing through the wire. If

the length of the solenoid became twice as long (2L), and all other quantities remained the same, the magnetic field inside the solenoid would _________.
Physics
1 answer:
wolverine [178]3 years ago
5 0

Answer:

The magnetic field inside the solenoid would decrease by a factor of 2.

Explanation:

The magnetic field, B, of a solenoid of length L, N windings, and radius b with a current, I, flowing through it is given as:

B = (N * r * I) / L

If the length of the solenoid is doubled, 2L,the magnetic field becomes:

B2 = (N * r * I) / 2L

B2 = ½ B

The magnetic field will decrease by a factor of 2.

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A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:
madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take g = 10\; \rm m \cdot s^{-2}.

a. The momentum of the ball would be approximately 60\;\rm kg \cdot m \cdot s^{-1} two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately \rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right) three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum p of an object is equal its mass m times its velocity v. That is: \vec{p} = m \cdot \vec{v}.

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. In other words, its velocity would become approximately 10\; \rm m \cdot s^{-1} more negative every second.

The initial velocity of the ball is 60\; \rm m \cdot s^{-1}. After two seconds, its velocity would have become 60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}. The momentum of the ball at that time would be around p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}.

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be 0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}. Accordingly, the ball's momentum at that moment would be p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right).

3 0
3 years ago
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iVinArrow [24]

Answer:

<em>The magnitude of the magnetic field will act in a direction towards me.</em>

<em></em>

Explanation:

When a charged particle enters a magnetic field, it is deflected. The direction of travel of the particle is deflected, but the kinetic energy of the particle is not affected. <em>The force experienced by a charged particle as it enters a magnetic field that acts perpendicular to the path of the velocity of the particle, will produce a force that is perpendicular to both the direction of travel of the particle and the direction of the magnetic field.</em> In this case, the proton moves in the y-direction, the magnetic field is in the x-direction, therefore the force experienced by the particle will be towards me.

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Answer:

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3 years ago
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