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Byju's Answer
Standard XII
Physics
Principle of Superposition
In an electro...
Question
In an electromagnetic wave, the amplitudes of magnetic field
H
o
and
E
o
in free space are related as
A
H
o
=
E
o
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B
H
=
E
o
c
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C
H
o
=
E
o
√
μ
o
ε
o
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D
H
o
=
E
o
√
ε
o
μ
o
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Solution
The correct option is
C
H
o
=
E
o
√
μ
o
ε
o
Relation between
E
o
and
H
o
E
o
H
o
=
1
C
=
√
μ
o
ε
o
⇒
H
o
=
E
o
√
μ
o
ε
o
Suggest Corrections
0
Similar questions
Q.
Pairs of electrodes and their corresponding standard electrode potentials are given. Arrange the cells constructed by these electrodes in increasing order of emf values.
(
a
)
E
l
e
c
t
r
o
d
e
→
A
,
E
o
=
−
2.92
V
and
E
l
e
c
t
r
o
d
e
→
B
,
E
o
=
−
2.71
V
(
b
)
E
l
e
c
t
r
o
d
e
→
C
,
E
o
=
−
0.760
V
and
E
l
e
c
t
r
o
d
e
→
D
,
E
o
=
−
0.44
V
(
c
)
E
l
e
c
t
r
o
d
e
→
B
,
E
o
=
−
2.71
V
and
E
l
e
c
t
r
o
d
e
→
D
,
E
o
=
−
0.44
V
(
d
)
E
l
e
c
t
r
o
d
e
→
A
,
E
o
=
−
2.92
V
and
E
l
e
c
t
r
o
d
e
→
C
,
E
o
=
−
0.76
V
Q.
The standard electrode potentials of the electrodes are given below. Arrange them in decreasing order of ease of oxidation:
(
a
)
E
l
e
c
t
r
o
d
e
−
I
,
E
o
=
−
2.89
V
(
b
)
E
l
e
c
t
r
o
d
e
−
I
I
,
E
o
=
−
0.16
V
(
c
)
E
l
e
c
t
r
o
d
e
−
I
I
I
,
E
o
=
0.77
V
(
d
)
E
l
e
c
t
r
o
d
e
−
I
V
,
E
o
=
−
2.93
V
(
e
)
E
l
e
c
t
r
o
d
e
−
V
,
E
o
=
−
1.67
V
Q.
For the reduction of
N
O
−
3
ion in an aqueous solution
E
o
is
0.96
V
. Values of
E
o
for some metal ions are given below:
V
2
+
(
a
q
)
+
2
e
−
→
V
;
E
o
=
−
1.19
V
F
e
3
+
(
a
q
)
+
3
e
−
→
F
e
;
E
o
=
−
0.04
V
A
u
3
+
(
a
q
)
+
3
e
−
→
A
u
;
E
o
=
1.40
V
H
g
2
+
(
a
q
)
+
2
e
−
→
H
g
;
E
o
=
+
0.86
V
The pair(s) of metals that is/are oxidised by
N
O
−
3
in aqueous solution is (are):
Q.
The value of E for elements are as follows:
F
e
+
2
+
2
e
→
F
e
;
E
o
=
0.44
V
F
e
+
3
+
1
e
→
F
e
+
2
;
E
o
=
+
0.77
V
S
n
+
4
+
2
e
→
S
n
+
2
;
E
o
=
+
0.13
V
I
2
+
2
e
→
2
I
;
E
o
=
+
0.54
V
C
r
3
+
+
3
e
→
C
r
;
E
o
=
0.74
V
Based on the above data correct statements are:
Q.
Based on
E
o
values given, state which statements are correct?
1
2
F
2
+
e
−
⟶
F
−
E
o
= 2.87 V
1
2
C
l
2
+
e
−
⟶
C
l
−
E
o
= 1.40 V
1
2
B
r
2
+
e
−
⟶
B
r
−
E
o
= 1.09 V
1
2
I
2
+
e
−
⟶
I
−
E
o
= 0.62 V
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