Answer:

The Magnitude of electric field is in the upward direction as shown directly towards the charge
.
Explanation:
Given:
- side of a square,

- charge on one corner of the square,

- charge on the remaining 3 corners of the square,

<u>Distance of the center from each corners</u>


∴Distance of center from corners, 
Now, electric field due to charges is given as:

<u>For charge
we have the field lines emerging out of the charge since it is positively charged:</u>

<u>Force by each of the charges at the remaining corners:</u>

<u> Now, net electric field in the vertical direction:</u>


<u>Now, net electric field in the horizontal direction:</u>


So the Magnitude of electric field is in the upward direction as shown directly towards the charge
.
The problem states that the distance travelled (d) is
directly proportional to the square of time (t^2), therefore we can write this in
the form of:
d = k t^2
where k is the constant of proportionality in furlongs /
s^2
<span>Using the 1st condition where d = 2 furlongs, t
= 2 s, we calculate for the value of k:</span>
2 = k (2)^2
k = 2 / 4
k = 0.5 furlongs / s^2
The equation becomes:
d = 0.5 t^2
Now solving for d when t = 4:
d = 0.5 (4)^2
d = 0.5 * 16
<span>d = 8 furlongs</span>
<span>
</span>
<span>It traveled 8 furlongs for the first 4.0 seconds.</span>
Answer:
I = 1.875 A
Explanation:
For this exercise we use Ampere's law
∫ B . ds = μ₀ I
We use a circular path around the wire whereby B and ds are parallel, whereby the dot product is reduced to the algebraic product
ds = 2π dr
B (2πr) = μ₀ I
I = B 2π R /μ₀
r= 7.5 cm = 0.075 m
calculate
I = (50 μ₀ /π) 2π 0.075 /μ₀
I = 1.875 A