Explanation:
n=50,r=0.02mn=50,r=0.02m,
I=5AandB=0.20TI=5AandB=0.20T
τismaxiμmwhensinθ=90∘τismaxiμmwhensinθ=90∘
τmax=niabsin90∘=mbτmax=niabsin90∘=mb
=50×5×3.14×4×10−4×2×10−1=50×5×3.14×4×10-4×2×10-1
=6.28×10−2Nm=6.28×10-2Nm
Given τ=12×τmaxτ=12×τmax
⇒sinsinθ=12⇒sinsinθ=12 or sinθ=30∘sinθ=30∘
=∠betweenareavar→randmag≠ticfield=∠betweenareavar→randmag≠ticfield.
So angle between magnetic field and the plane of the coil
=90∘−30∘=60∘=90∘-30∘=60∘.
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Answer:
the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²
Explanation:
Given the data in the question;
To determine the maximum intensity of an electromagnetic wave, we use the formula;
= 
ε₀cE
²
where ε₀ is permittivity of free space ( 8.85 × 10⁻¹² C²/N.m² )
c is the speed of light ( 3 × 10⁸ m/s )
E is the maximum magnitude of the electric field
first we calculate the maximum magnitude of the electric field ( E )
E = 350/f kV/m
given that frequency of 60 Hz, we substitute
E = 350/60 kV/m
E = 5.83333 kV/m
E = 5.83333 kV/m × ( 
)
E = 5833.33 N/C
so we substitute all our values into the formula for intensity of an electromagnetic wave;
= ε₀cE²
= × ( 8.85 × 10⁻¹² C²/N.m² ) × ( 3 × 10⁸ m/s ) × ( 5833.33 N/C )²
= 45 × 10³ W/m²
= 45 × 10³ W/m² × ( 
)
= 45 kW/m²
Therefore, the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²
Answer:
No
Explanation:
You could try to give it enough to fill all valence electrons in all of the atoms in the conductor, but practically this could not be achieved.
Option (d) is correct.Fission and fusion convert nuclear energy to both radiant and thermal energy.
fission is a process in which bigger nucleus breaks into two or more smaller nuclei with the liberation of a large amount of energy.
Fusion is a process in which two small nuclei fuse together to from an intermediate size nucleus with the liberation of tremendous amount of energy
During fission and fusion, energy is released in the form of both light (radiation) and heat.we can then convert that energy into useful form.
