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Test Index
Magnetism and Matter
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Section:
Physics
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© examsnet.com
Question : 63 of 75
Marks:
+1
,
-0
A fighter plane of length 20 m, wing span (distance from tip of one wing to the tip of the other wing) of 15 m and height 5 m is flying towards east over Delhi. Its speed is
240
m/s
240\,\text{m/s}
240
m/s
.The earth's magnetic field over Delhi is
5
×
1
0
−
5
T
5\times10^{-5}\,\text{T}
5
×
1
0
−
5
T
with the declination angle
∼
0
∘
\sim0^{\circ}
∼
0
∘
and dip of θ such that
sin
θ
=
2
3
\sin\theta=\frac{2}{3}
sin
θ
=
3
2
.If the voltage developed is
V
B
V_B
V
B
between the lower and upper side of the plane and VW between the tips of the wings then
V
B
V_B
V
B
and
V
W
V_W
V
W
are close to :
[Main 10 Apr 2016]
V
B
=
45
mV
;
V
W
=
120
mV
V_B=45\,\text{mV};V_W=120\,\text{mV}
V
B
=
45
mV
;
V
W
=
120
mV
with right side of pilot at higher voltage
V
B
=
45
mV
;
V
W
=
120
mV
V_B=45\,\text{mV};V_W=120\,\text{mV}
V
B
=
45
mV
;
V
W
=
120
mV
with left side of pilot at higher voltage
V
B
=
40
mV
;
V
W
=
135
mV
V_B=40\,\text{mV};V_W=135\,\text{mV}
V
B
=
40
mV
;
V
W
=
135
mV
with right side of pilot at high voltage
V
B
=
40
mV
;
V
W
=
135
mV
V_B=40\,\text{mV};V_W=135\,\text{mV}
V
B
=
40
mV
;
V
W
=
135
mV
with left side of pilot at higher voltage
Validate
Solution:
V
B
=
B
4
(
5
)
(
240
)
V_B=B_4(5)(240)
V
B
=
B
4
(
5
)
(
240
)
B
H
=
B
cos
θ
B_H=B\cos\theta
B
H
=
B
cos
θ
B
H
=
5
5
×
1
0
−
5
3
B_H=\frac{5\sqrt{5}\times10^{-5}}{3}
B
H
=
3
5
5
×
1
0
−
5
B
v
=
10
3
×
1
0
−
5
B_v=\frac{10}{3}\times10^{-5}
B
v
=
3
10
×
1
0
−
5
T
V
B
=
5
5
3
×
1
0
−
5
×
5
×
240
V_B=\frac{5\sqrt{5}}{3}\times10^{-5}\times5\times240
V
B
=
3
5
5
×
1
0
−
5
×
5
×
240
V
B
=
44.6
mV
=
45
mV
V_B=44.6\,\text{mV}=45\,\text{mV}
V
B
=
44.6
mV
=
45
mV
v
w
=
B
v
l
V
v_w=B_vlV
v
w
=
B
v
l
V
=
1
0
−
4
×
1200
=10^{-4}\times1200
=
1
0
−
4
×
1200
V
ω
=
120
V_\omega=120
V
ω
=
120
mV
(left side at fighter voltage)
© examsnet.com
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