A copper wire has a circular cross section with a radius of 1.00 mm. (a) If the wire carries a current of 2.80 A, find the drift
speed of the electrons in the wire. (Assume the density of charge carriers (electrons) in a copper wire is n = 8.46 1028 electrons/m3.) 2.1e-6 Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s (b) All other things being equal, what happens to the drift speed in wires made of metal having a larger number of conduction electrons per atom than copper? Explain
1 answer:
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Question:
A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.
(a)60 (b)90 (c)120
Answer:
(a)5.42 N (b)6.26 N (c)5.42 N
Explanation:
From the question
Length of wire (L) = 2.80 m
Current in wire (I) = 5.20 A
Magnetic field (B) = 0.430 T
Angle are different in each part.
The magnetic force is given by
So from data
Now sub parts
(a)
(b)
(c)
Answer:
Because there is no air resistance
Explanation:
When an object falls on Earth, there are essentially two forces acting on it:
- The force of gravity, downward, equal to the weight of the object:
where m is the mass and g the acceleration due to gravity
- The air resistance, F, which acts upward, and whose magnitude is generally proportional to v, the speed of the object
When the object starts its fall, its initial speed is zero: v = 0, so the air resistance is also zero: F=0, and the object accelerates downward due to gravity.
However, as it accelerates downward, its speed increases, and so does the air resistance F. However, F is upward, opposite to the direction of motion, therefore it reduces the net acceleration of the object; at a certain point, the magnitude of the air resistance will become equal to the weight, so that
mg = F
and at that point, the net acceleration of the object will become zero: this means that the object will continue its fall at a constant velocity, called terminal velocity.
On the Moon instead, there is no air resistance: this means that for an object falling down, the speed keeps increasing due to the effect of gravity, and it will never reach a terminal value: therefore, the final velocity at the impact will be much higher than on the Earth, if we assume the two objects have been dropped from a very high altitude from the surface.
Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:
where F is the force applied, and
is the stretch of the spring with respect to its equilibrium position. Using the data, we find
Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
where F is the force applied, and
Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring: