this is an equation that you need to solve for motional emf. motional emf=vBL, where v is velocity in meters/second, B is magnetic field in Teslas and L is length or distance the rails are apart from each other. when we plug everything into the formula given above, we get: motional emf=5m/s*0.80T*0.20m. solving all this we get 0.8 volts. pretty sure that since they are giving you the direction of the field, they want to know which way the current will flow . since the conductor is moving from left to right the area of the field is increasing which means magnetic flux is increasing as Ф(magnetic flux)=B(magnetic field)*A(area)*cosФ(little phi is the angle to the normal. in this case little fee is 0 degrees so the cosФ doesn't matter). so ↑Ф=B↑A. if magnetic flux is increasing, the induced magnetic field is in the opposite direction as the original magnetic field meaning the induced magnetic field will be out of the page. using the right hand rule which says that if the field is in to the page, the current should go clockwise and if the field is out of the page, the current is counterclockwise so that means that the current should be going counter clockwise since the induced field is going out of the screen. the top of the conducting wire will have its current go to the left and the bottom of the conducting wire will have the current go to the right.
Answer:
A
Explanation:
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weight less on moon than on earth.
high on lift off - G force
low in orbit.
zero at a point between earth and moon
Answer:
The electric field at origin is 3600 N/C
Solution:
As per the question:
Charge density of rod 1, 
Charge density of rod 2, 
Now,
To calculate the electric field at origin:
We know that the electric field due to a long rod is given by:
Also,
(1)
where
K = electrostatic constant = 
R = Distance
= linear charge density
Now,
In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.
At x = - 1 cm = - 0.01 m:
Using eqn (1):
(towards)
Now, at x = 1 cm = 0.01 m :
Using eqn (1):
(towards)
Now, the total field at the origin is the sum of both the fields:
