The short vertical parts adjacent to it also reach into the magnetic field and should experience forces. why can we neglect them
1 answer:
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Answer:
Final speed of the crate is 15 m/s
Explanation:
As we know that constant force F = 80 N is applied on the object for t = 12 s
Now we can use definition of force to find the speed after t = 12 s
so here we know that object is at rest initially so we have
Now for next 6 s the force decreases to ZERO linearly
so we can write the force equation as
now again by same equation we have
put t = 6 s
Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by
where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find: