The intensity of the magnetic force exerted on the wire due to the presence of the magnetic field is given by
where
I is the current in the wire
L is the length of the wire
B is the magnetic field intensity
is the angle between the direction of the wire and the magnetic field
In our problem, L=65 cm=0.65 m, I=0.35 A and B=1.24 T. The force on the wire is F=0.26 N, therefore we can rearrange the equation to find the sine of the angle:
and so, the angle is
Answer:
Explanation:
Electron's kinetic energy = 2 eV
= 2 x 1.6 x 10⁻¹⁹ J
1/2 m v² = 3.2 x 10⁻¹⁹
1/2 x 9.1 x 10⁻³¹ x v² = 3.2 x 10⁻¹⁹
v² = .703 x 10¹²
v = .8385 x 10⁶ m/s
Electrons revolve in a circular orbit when forced to travel in a magnetic field whose radius can be expressed as follows
r = mv / Bq
where m , v and q are mass , velocity and charge of electron .
here given magnetic field B = 90 mT
= 90 x 10⁻³ T
Putting these values in the expression above
r = mv / Bq
= 
= .052 mm.
Potential Energy= 24m * 14kg * 9.8N/kg = 3292.8J
We have the meats Arby’s we beat them kids
Answer:
ΔE> E_minimo
We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference
Explanation:
For this exercise let's use the electric field expression
E = k q / r²
where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee
let's calculate the field for each charge
Q = 24 pC = 24 10⁻¹² C
E₁ = 9 10⁹ 24 10⁻¹² / 0.20²
E₁ = 5.4 N / C
Q = 32 pC = 32 10⁻¹² C
E₂ = 9 10⁹ 32 10⁻¹² / 0.2²
E₂ = 7.2 N / C
let's find the difference between these two fields
ΔE = E₂ -E₁
ΔE = 7.2 - 5.4
ΔE = 1.8 N / C
the minimum detection field is
E_minimum = 0.77 N / C
ΔE> E_minimo
We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference
