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Test Index
AIEEE 2009 Solved Paper
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Section:
Physics
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© examsnet.com
Question : 49 of 89
Marks:
+1
,
-0
A current loop
A
B
C
D
A B C D
A
BC
D
is held fixed on the plane of the paper as shown in the figure. The arcs
B
C
B C
BC
(radius
=
b
=b
=
b
) and
D
A
D A
D
A
(radius
=
a
=a
=
a
) of the loop are joined by two straight wires
A
B
A B
A
B
and
C
D
C D
C
D
. A steady current
I
I
I
is flowing in the loop. Angle made by
A
B
A B
A
B
and
C
D
C D
C
D
at the origin
O
O
O
is
3
0
∘
30^{\circ}
3
0
∘
. Another straight thin wire steady current
I
1
I_1
I
1
flowing out of the plane of the paper is kept at the origin.
Due to the presence of the current
I
1
I_1
I
1
at the origin:
[AIEEE 2009]
The forces on
A
D
A D
A
D
are
B
C
B C
BC
are zero.
The magnitude of the net force on the loop is given by
I
1
I
4
π
μ
0
[
2
(
b
−
a
)
+
π
3
(
a
+
b
)
]
.
\; \frac{I_1 I}{4\pi} \mu_0 \left[2(b-a) + \frac{\pi}{3}(a+b)\right] \; \text{. } \;
4
π
I
1
I
μ
0
[
2
(
b
−
a
)
+
3
π
(
a
+
b
)
]
.
The magnitude of the net force on the loop is given by
μ
0
I
I
1
24
a
b
(
b
−
a
)
\; \frac{\mu_0 I I_1}{24ab}(b-a)
24
ab
μ
0
I
I
1
(
b
−
a
)
.
The forces on
A
B
A B
A
B
and
D
C
D C
D
C
are zero.
Validate
Solution:
F
→
=
I
(
ℓ
→
×
B
→
)
\overset{\rightarrow}{F}=I\left(\overset{\rightarrow}{\ell}\times \overset{\rightarrow}{B}\right)
F
→
=
I
(
ℓ
→
×
B
→
)
The force on
A
D
A D
A
D
and
B
C
B C
BC
due to current
I
1
I_1
I
1
is zero. This is because the directions of current element
I
d
ℓ
→
\overset{\rightarrow}{I d\ell}
I
d
ℓ
→
and magnetic field
B
→
\overset{\rightarrow}{B}
B
→
are parallel.
© examsnet.com
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