The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The correct answer is option c
Explanation:
Faraday states that when there is a change in magnetic field of a coil of a wire, it means that there exist an emf in the circuit which in induced due to the change in the magnetic flux
From the question two separate but nearby coils are mounted along the axis. First coil is connected to the power supply and the current flow is controlled by the supply.When the current alternates, it would produce magnetic field ,also the second coil is connected to an ammeter which indicates the current that is flowing in it when current in the first coil changes
This magnetic field that is produce would cause a change flux which would induce current in the second coil so the ammeter would indicate current flow in the second coil
a is incorrect because the current in fir coil is not change hence flux won't change therefore current is is not induced in second coil
This is the same reason b is incorrect
d is incorrect due to the fact that when the second coil is connected to a power supply by rewiring it to be in series with first coil the law of electromagnetism would no longer hold so he ammeter would show no reading
Answer:
The load has a mass of 2636.8 kg
Explanation:
Step 1 : Data given
Mass of the truck = 7100 kg
Angle = 15°
velocity = 15m/s
Acceleration = 1.5 m/s²
Mass of truck = m1 kg
Mass of load = m2 kg
Thrust from engine = T
Step 2:
⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:
T = (m1+m2)*g*sinθ
⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes m1*gsinθ .
Resultant force on truck is F = T – m1*gsinθ
F causes the acceleration of the truck: F= m*a
This gives the equation:
T – m1*gsinθ = m1*a
T = m1(a + gsinθ)
Combining both equations gives:
(m1+m2)*g*sinθ = m1*(a + gsinθ)
m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ
m2*g*sinθ = m1*a
Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:
m2*g*sinθ = (7100 – m2)*a
m2*g*sinθ = 7100a – m2a
m2*gsinθ + m2*a = 7100a
m2* (gsinθ + a) = 7100a
m2 = 7100a/(gsinθ + a)
m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)
m2 = 2636.8 kg
The load has a mass of 2636.8 kg
Answer:
V0=27.4 m/s; t=0.8 s
Explanation:
Final position y=37.0 m, time = 2.3 s; Initial position is set to be zero. We calculate the initial speed with the kinematics equation:
We solve for initial speed

Now, using the same expression we estimated time to first reach 18.5 m :
Second order equation with solutions
t1=0.8 s and t2=4.8 s
The first time corresponds to the first reach.
Answer:
a) E=228391.8 N/C
b) E=-59345.91N/C
Explanation:
You can use Gauss law to find the net electric field produced by both line of charges.

Where E1 and E2 are the electric field generated at a distance of r1 and r2 respectively from the line of charges.
The net electric field at point r will be:

a) for y=0.200m, r1=0.200m and r2=0.200m:
![E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.200m}-\frac{2.26*10^{-6}C}{0.200m}}]=228391.8N/C](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%5Cpi%288.85%2A10%5E%7B-12%7DC%5E2%2FNm%5E2%29%7D%5B%5Cfrac%7B4.80%2A10%5E%7B-6%7DC%7D%7B0.200m%7D-%5Cfrac%7B2.26%2A10%5E%7B-6%7DC%7D%7B0.200m%7D%7D%5D%3D228391.8N%2FC)
b) for y=0.600m, r1=0.600m, r2=0.200m:
![E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.600m}-\frac{2.26*10^{-6}C}{0.200m}}]=-59345.91N/C](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%5Cpi%288.85%2A10%5E%7B-12%7DC%5E2%2FNm%5E2%29%7D%5B%5Cfrac%7B4.80%2A10%5E%7B-6%7DC%7D%7B0.600m%7D-%5Cfrac%7B2.26%2A10%5E%7B-6%7DC%7D%7B0.200m%7D%7D%5D%3D-59345.91N%2FC)