Answer:
the magnitude of the electric field is 1.25 N/C
Explanation:
The induced emf in the cube ε = LB.v where B = magnitude of electric field = 5 T , L = length of side of cube = 1 cm = 0.01 m and v = velocity of cube = 1 m/s
ε = LB.v = 0.01 m × 5 T × 1 m/s = 0.05 V
Also, induced emf in the cube ε = ∫E.ds around the loop of the cube where E = electric field in metal cube
ε = ∫E.ds
ε = Eds since E is always parallel to the side of the cube
= E∫ds ∫ds = 4L since we have 4 sides
= E(4L)
= 4EL
So,4EL = 0.05 V
E = 0.05 V/4L
= 0.05 V/(4 × 0.01 m)
= 0.05 V/0.04 m
= 1.25 V/m
= 1.25 N/C
So, the magnitude of the electric field is 1.25 N/C
<span>For this particular problem, we use Ohm's Law. This law deals with the relation between
voltage and current in an ideal conductor. It states that: Potential difference
across a conductor is proportional to the current that pass through it. It is
expressed as V=IR. Using the equation, we can isolate I or the current to one side and the other terms to the other side. We calculate as follows:
V = IR
I = V/R
I = 12 V / 20 </span><span>Ω
I = 0.6 amperes
Therefore, the current that is flowing through the wire supplied with 12 V and having a resistance of 20 </span><span>Ω would be 0.6 amperes.</span>
Earth quake cause the plates move to Cause either a big or small reaction called an earth quake
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