Answer:
The length of the wire is 579 m
Explanation:
Given;
current on the wire, I = 11.3-mA
magnetic field of the wire, B = (16.2i + 2.4 ĵ) T
Magnitude of force experience by the wire, F = 15.7 N
Magnitude of force experience by current carrying wire at a given a magnetic field strength is calculated as;
F = BILsinθ
Where;
B is magnitude of magnetic field
F is the force on the wire
L is length of the wire
θ is direction of the magnetic field
Length of the wire is calculated as;
Therefore, the length of the wire is 579 m
Answer:
minimum angle is 128.69°
Explanation:
given data
player velocity with respect ground v1 = 3.5 m/s
ball velocity with respect himself v2 = 5.6 m/s
to find out
smallest angle
solution
we know ball velocity with respect field will be
ball velocity = v1 +v2
ball velocity = 3.5 + 5.6 = 9.1m/s
we consider angle that player hit ball is θ
then by as per figure triangle
cosθ =
cosθ =
θ = 51.31
so minimum angle is 180 - 51.31 = 128.69°
Answer:
a. cosθ b. E.A
Explanation:
a.The electric flux, Φ passing through a given area is directly proportional to the number of electric field , E, the area it passes through A and the cosine of the angle between E and A. So, if we have a surface, S of surface area A and an area vector dA normal to the surface S and electric field lines of field strength E passing through it, the component of the electric field in the direction of the area vector produces the electric flux through the area. If θ the angle between the electric field E and the area vector dA is zero ,that is θ = 0, the flux through the area is maximum. If θ = 90 (perpendicular) the flux is zero. If θ = 180 the flux is negative. Also, as A or E increase or decrease, the electric flux increases or decreases respectively. From our trigonometric functions, we know that 0 ≤ cos θ ≤ 1 for 90 ≤ θ ≤ 0 and -1 ≤ cos θ ≤ 0 for 180 ≤ θ ≤ 90. Since these satisfy the limiting conditions for the values of our electric flux, then cos θ is the required trigonometric function. In the attachment, there is a graph which shows the relationship between electric flux and the angle between the electric field lines and the area. It is a cosine function
b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A
