Answer:
Explanation:
Given the electric field
E=(2.5• j + 3.5• k) ×10³ N/C
Given the radius of the circular path r=2.5m.
Φ=?
Flux In an electric field is given as
Φ=∮E•dA. From r=0 to r=2.5m
Given that A of a striker is
The area is in yz plane then, it's normal will be in x-direction
A=πr²
Then, dA=2πrdr •i
Φ=∮E•dA. From r=0 to r=2.5m
Φ=∮(2.5j + 3.5k)×10³•(2πrdr i)From r=0 to r=2.5m
Note that, i•i=j•j=k•k=1
i•j=j•i=k•i=i•k=j•k=k•j=0
Then,
Φ=10³∮ 0dr From r=0 to r=2.5m
Φ= 0 Nm²/C
b. When it is directed in the angle of 45° between above xy
Then,
dA=(2πrCos45 i + 2πrSin45 j) dr
Φ=∮E•dA. From r=0 to r=2.5m
Φ=∮(2.5j + 3.5k)×10³•(2πrCos45dr i + 2πrSin45 j). From r=0 to r=2.5m.
Φ=10³∮(2.5j + 3.5k)•(2πrCos45dr i + 2πrSin45 j). From r=0 to r=2.5m.
Φ=10³∮ 5πrSin45 dr
Φ=10³×5π×Sin45 [r²/2] r=0 r=2.5m
Φ=10³ ×5π×Sin45×½[2.5²-0²]
Φ=10³× 5π × Sin45 × ½ × 6.25
Φ=10³×34.71
Φ=34.71×10³ Nm²/C