A straight, vertical wire carries a current of 2.15 A downward in a region between the poles of a large superconducting electrom
1 answer:
Answer:
a) F = 0.01234N to the south direction
b) F = 0.01234N to the north direction
c) F = 0.01234N
d) North of West
Explanation:
The magnetic force in a magnetic field is given by:
Length of the wire, l = 1 cm = 0.01 m
Current flowing through the wire, I = 2.15 A
a) If the magnetic field direction is east
∅ = 90
F = 0.574 * 2.15 * 0.01
F = 0.01234 N to the south direction according to the Fleming's Right Hand Rule
b) If the magnetic field direction is south
The magnitude of the magnetic force remains the same
That is , F = BIL sin 90
F = 0.01234 N to the west
c) If the magnetic field direction is 30 degrees south
The angle between the magnetic field and the length of the wire still remains 90 degrees
Therefore the magnitude of the force still remains 0.01234
F = 0.01234 N
d) the direction of the force is the North of West
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Answer:
Option 4. 0.05 J
Solution:
As per the question:
Spring constant, k = 10 N/m
Equilibrium position, x = 0.1 m
Now,
The potential energy of the spring is given by:
And also from the principle of conservation of energy:
KE = U (1)
where
KE = Maximum Kinetic Energy
U = Potential energy
Thus
KE = U =
KE = U =