A current carrying wire of length 50cm is positioned perpendicular to a uniform magnetic field
1 answer:
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Yield strength of mild steel is given as
here we know that
now we can rearrange it as
here we know that
so maximum applied force will be F = 1.23 * 10^3 N
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 =