A kangaroo jumps up with an initial velocity of
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Strength of the magnetic field: C) 20 T
Explanation:
When a conductive wire moves in a region with magnetic field, an electromotive force is induced in the wire due to the phenomenon of electromagnetic induction.
If the wire moves perpendicular to the field, the magnitude of the induced emf in the wire is given by
where
B is the strength of the magnetic field
L is the length of the wire
v is the speed of the wire
For the wire in this problem, we have:
is the emf induced in the wire
v = 3.0 m/s is the speed of the wire
L = 0.20 m is its length
Solving for B, we find the strength of the magnetic field:
Learn more about magnetic fields:
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Answer:
A) x and y components of the electric field (Ep) at the origin.
Epx = -1620.5 N/C
Epy = -1620.5 N/C
Ep = (1620.5(-i)+1620.5(-j)) N/C
B) Magnitude of the electric field (Ep) at the origin.
Ep= 2291.7 N/C
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
Equivalence
1nC= 10⁻⁹C
Data
K= 9x10⁹N*m²/C²
q₁ = q₂= +6.5nC=+6.5 *10⁻⁹C
d₁=d₂=0.190m
Graphic attached
The attached graph shows the field due to the charges:
Ep₁ : Electric Field at point P (x=0, y=0) due to charge q₁. As the charge q₁ is positive (q₁+) ,the field leaves the charge.
Ep₂: Electric Field at point P (x=0, y=0) due to charge q₂. As the charge q₂ is positive (q₂+) ,the field leaves the charge
Ep: Total field at point P due to charges q₁ and q₂.
Because q₁ = q₂ and d₁ = d₂, then, the magnitude of Ep₁ is equal to the magnitude of Ep₂
Ep₁ = Ep₂ = k*q/d² = 9*10⁹*6.5*10⁻⁹/0.190m² = 1620.5 N/C
Look at the attached graphic :
Epx = Ep₁= -1620.5 N/C
Epy = Ep₂= -1620.5 N/C
A) x and y components of the electric field (Ep) at the origin.
Ep = (1620.5(-i)+1620.5(-j)) N/C
B) Magnitude of the electric field (Ep) at the origin.
