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Curl of a Vector Field
If H =2yâx +z...
Question
If
→
H
=
2
y
^
a
x
+
(
z
2
−
x
2
)
^
a
y
+
3
y
^
a
z
, then the curl of vector field
→
H
at origin is
A
√
12
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B
√
13
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C
√
14
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D
√
15
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Solution
The correct option is
B
√
13
▽
×
→
H
=
∣
∣ ∣ ∣
∣
i
j
k
∂
∂
x
∂
∂
y
∂
∂
z
2
y
(
z
2
−
x
2
)
3
y
∣
∣ ∣ ∣
∣
=
^
i
(
3
−
2
z
)
+
^
k
(
−
2
x
−
2
)
=
(
3
−
2
z
)
^
i
−
(
2
x
+
2
)
^
k
At the origin , x =0, z= 0
▽
×
→
H
=
3
^
i
−
2
^
k
|
▽
×
→
H
| =
√
3
2
+
2
2
=
√
13
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Given the general Vector
→
A
=
(
s
i
n
2
ϕ
)
^
a
ϕ
in cylindrical system , then the curl of vector
→
A
at (2,
π
4
,0) is
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