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JEE Mains 05 Sep 2020 Shift 2 Solved Paper
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Section:
Physics
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© examsnet.com
Question : 5 of 75
Marks:
+1
,
-0
An infinitely long straight wire carrying current I, one side opened rectangular loop and a conductor
C
C
C
with a sliding connector are located in the same plane, as shown in the figure. The connector has length
l
l
l
and resistance
R
R
R
. It slides to the right with a velocity v. The resistance of the conductor and the self inductance of the loop are negligible. The induced current in the loop, as a function of separation
r
,
r,
r
,
between the connector and the straight wire is :
[5 Sep 2020 Shift 2]
μ
0
Ï€
I
v
l
R
r
\frac{\mu_{0}}{\pi} \frac{I v l}{R r}
Ï€
μ
0
​
​
R
r
I
v
l
​
μ
0
2
Ï€
I
v
l
R
r
\frac{\mu_{0}}{2\pi} \frac{I v l}{R r}
2
Ï€
μ
0
​
​
R
r
I
v
l
​
2
μ
0
Ï€
I
v
l
R
r
\frac{2\mu_{0}}{\pi} \frac{I v l}{R r}
Ï€
2
μ
0
​
​
R
r
I
v
l
​
μ
0
4
Ï€
I
v
l
R
r
\frac{\mu_{0}}{4\pi} \frac{I v l}{R r}
4
Ï€
μ
0
​
​
R
r
I
v
l
​
Validate
Solution:
B
=
μ
0
i
2
Ï€
r
B = \frac{\mu_{0} i}{2 \pi r}
B
=
2
Ï€
r
μ
0
​
i
​
Ï•
=
μ
0
i
2
Ï€
r
l
d
r
\phi = \frac{\mu_{0} i}{2 \pi r} l dr
Ï•
=
2
Ï€
r
μ
0
​
i
​
l
d
r
⇒
d
Ï•
d
t
=
μ
0
i
l
2
Ï€
r
â‹…
d
r
d
t
\Rightarrow \frac{d\phi}{dt} = \frac{\mu_{0} i l}{2 \pi r} \cdot \frac{dr}{dt}
⇒
d
t
d
Ï•
​
=
2
Ï€
r
μ
0
​
i
l
​
â‹…
d
t
d
r
​
⇒
e
=
μ
0
2
Ï€
â‹…
i
v
l
r
\Rightarrow e = \frac{\mu_{0}}{2 \pi} \cdot \frac{i v l}{r}
⇒
e
=
2
Ï€
μ
0
​
​
â‹…
r
i
v
l
​
i
=
e
R
=
μ
0
2
Ï€
â‹…
i
v
l
R
r
i = \frac{e}{R} = \frac{\mu_{0}}{2 \pi} \cdot \frac{i v l}{R r}
i
=
R
e
​
=
2
Ï€
μ
0
​
​
â‹…
R
r
i
v
l
​
© examsnet.com
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