Choose the correct statement about the acceleration of the car.
1 answer:
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Complete question:
Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.
Answer:
The magnitude of this field is 826 N/C
Explanation:
Given;
The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m
PEsinθ = T
where;
E is the magnitude of the electric field
P is the dipole moment
First, we determine the magnitude dipole moment;
Magnitude of dipole moment = q*r
P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m
Finally, we determine the magnitude of this field;
E = 826 N/C (in three significant figures)
Therefore, the magnitude of this field is 826 N/C
Answer:
Explanation:
Given the following :
Speed (V) = speed of 2.30×10^7 m/s
Acceleration (a) = 1.70×10^13 m/s^2
Using the right hand rule provided by Lorentz law:
B = F / qvSinΘ
Where B = magnitude of the magnetic field
v = speed of the particle
Θ = 90° (perpendicular to the field)
q = charge of the particle
SinΘ = sin90° = 1
Note F = ma
Therefore,
B = ma / qvSinΘ
Mass of proton = 1.67 × 10^-27
Charge = 1.6 × 10^-19 C
B = [(1.67 × 10^-27) × (1.70 × 10^13)] / (1.6 × 10^-19) × (2.30 × 10^7) × 1
B = 2.839 × 10^-14 / 3.68 × 10^-12
B = 0.7715 × 10^-2
B = 7.72 × 10^-3 T
2) Magnetic field will be in the negative y direction according to the right hand thumb rule.
Since Velocity is in the positive z- direction, acceleration in the positive x - direction, then magnetic field must be in the negative y-direction.
