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A long solenoid of diameter 0.1 m has 2 Ć 10 4 2 \times 10^4 2 Ć 1 0 4 turns per meter. At the center of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is 10 Ļ 2 Ī© , 10 \pi^2 \Omega, 10 Ļ 2 Ī© , the total charge flowing through the coil during this time is
hard
Moving Charges and Magnetism
2017
physics
16 Ļ Ī¼ C 16 \pi \mu C 16 Ļ Ī¼ C
32 Ļ Ī¼ C 32 \pi \mu C 32 Ļ Ī¼ C
Explanation To solve this problem, we need to calculate the total charge flowing through the coil due to the changing magnetic field in the solenoid. \\
Given: \\
ā¢ Diameter of the solenoid = 0.1 ā = 0.1 \, = 0.1 m ā \Rightarrow ā Radius r s = 0.05 ā r_s = 0.05 \, r s ā = 0.05
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m
ā¢ Turns per meter of the solenoid
= 2 Ć 10 4 = 2 \times 10^4 \\
= 2 Ć 1 0 4 ā¢ Initial current in the solenoid
I i = 4 ā I_i = 4 \, I i ā = 4 A
ā¢ Final current in the solenoid
I f = 0 ā I_f = 0 \, I f ā = 0 A
ā¢ Time interval
Ī t = 0.05 ā \Delta t = 0.05 \, Ī t = 0.05 s
ā¢ Number of turns in the coil
N c = 100 N_c = 100 \\
N c ā = 100 ā¢ Radius of the coil
r c = 0.01 ā r_c = 0.01 \, r c ā = 0.01 m
ā¢ Resistance of the coil
R = 10 Ļ 2 ā Ī© R = 10 \pi^2 \, \Omega \\
R = 10 Ļ 2 Ī© The magnetic field inside the solenoid is given by:
B = Ī¼ 0 n I \\
B = \mu_0 n I \\
B = Ī¼ 0 ā n I where
Ī¼ 0 = 4 Ļ Ć 10 ā 7 ā \mu_0 = 4\pi \times 10^{-7} \, Ī¼ 0 ā = 4 Ļ Ć 1 0 ā 7 T m/A
is the permeability of free space,
n = 2 Ć 10 4 ā \\
n = 2 \times 10^4 \, n = 2 Ć 1 0 4 turns/m
is the number of turns per meter, and
is the current.
The change in magnetic field
is:
Ī B = Ī¼ 0 n ( I f ā I i ) = Ī¼ 0 n ( 0 ā 4 ) = ā 4 Ī¼ 0 n \\
\Delta B = \mu_0 n (I_f - I_i) = \mu_0 n (0 - 4) = -4 \mu_0 n \\
Ī B = Ī¼ 0 ā n ( I f ā ā I i ā ) = Ī¼ 0 ā n ( 0 ā 4 ) = ā 4 Ī¼ 0 ā n The magnetic flux through the coil is given by:
Ī¦ = B ā
A \\
\Phi = B \cdot A \\
Ī¦ = B ā
A where
A = Ļ r c 2 A = \pi r_c^2 A = Ļ r c 2 ā is the area of the coil.
The change in magnetic flux
is:
Ī Ī¦ = Ī B ā
A = ( ā 4 Ī¼ 0 n ) ā
( Ļ r c 2 ) \\
\Delta \Phi = \Delta B \cdot A = (-4 \mu_0 n) \cdot (\pi r_c^2) \\
ĪĪ¦ = Ī B ā
A = ( ā 4 Ī¼ 0 ā n ) ā
( Ļ r c 2 ā ) The induced emf
in the coil is given by Faraday's law:
E = ā N c Ī Ī¦ Ī t \\
\mathcal{E} = -N_c \frac{\Delta \Phi}{\Delta t} \\
E = ā N c ā Ī t ĪĪ¦ ā Substitute the values:
E = ā 100 ā
( ā 4 ā
4 Ļ Ć 10 ā 7 ā
2 Ć 10 4 ā
Ļ ā
( 0.01 ) 2 ) 0.05 \\
\mathcal{E} = -100 \cdot \frac{(-4 \cdot 4\pi \times 10^{-7} \cdot 2 \times 10^4 \cdot \pi \cdot (0.01)^2)}{0.05} \\
E = ā 100 ā
0.05 ( ā 4 ā
4 Ļ Ć 1 0 ā 7 ā
2 Ć 1 0 4 ā
Ļ ā
( 0.01 ) 2 ) ā Simplify:
E = 100 ā
4 ā
4 Ļ Ć 10 ā 7 ā
2 Ć 10 4 ā
Ļ ā
10 ā 4 0.05 E = 100 ā
32 Ļ 2 Ć 10 ā 7 ā
2 Ć 10 4 ā
10 ā 4 0.05 E = 100 ā
64 Ļ 2 Ć 10 ā 7 ā
10 ā 4 0.05 E = 100 ā
64 Ļ 2 Ć 10 ā 11 0.05 E = 100 ā
128 Ļ 2 Ć 10 ā 11 E = 128 Ļ 2 Ć 10 ā 9 ā \\
\mathcal{E} = 100 \cdot \frac{4 \cdot 4\pi \times 10^{-7} \cdot 2 \times 10^4 \cdot \pi \cdot 10^{-4}}{0.05} \\
\mathcal{E} = 100 \cdot \frac{32 \pi^2 \times 10^{-7} \cdot 2 \times 10^4 \cdot 10^{-4}}{0.05} \\
\mathcal{E} = 100 \cdot \frac{64 \pi^2 \times 10^{-7} \cdot 10^{-4}}{0.05} \\
\mathcal{E} = 100 \cdot \frac{64 \pi^2 \times 10^{-11}}{0.05} \\
\mathcal{E} = 100 \cdot 128 \pi^2 \times 10^{-11} \\
\mathcal{E} = 128 \pi^2 \times 10^{-9} \, E = 100 ā
0.05 4 ā
4 Ļ Ć 1 0 ā 7 ā
2 Ć 1 0 4 ā
Ļ ā
1 0 ā 4 ā E = 100 ā
0.05 32 Ļ 2 Ć 1 0 ā 7 ā
2 Ć 1 0 4 ā
1 0 ā 4 ā E = 100 ā
0.05 64 Ļ 2 Ć 1 0 ā 7 ā
1 0 ā 4 ā E = 100 ā
0.05 64 Ļ 2 Ć 1 0 ā 11 ā E = 100 ā
128 Ļ 2 Ć 1 0 ā 11 E = 128 Ļ 2 Ć 1 0 ā 9 V
The total charge
flowing through the coil is given by:
Q = E ā
Ī t R \\
Q = \frac{\mathcal{E} \cdot \Delta t}{R} \\
Q = R E ā
Ī t ā Substitute the values:
Q = 128 Ļ 2 Ć 10 ā 9 ā
0.05 10 Ļ 2 Q = 128 Ć 0.05 Ć 10 ā 9 10 Q = 6.4 Ć 10 ā 9 10 Q = 6.4 Ć 10 ā 10 ā \\
Q = \frac{128 \pi^2 \times 10^{-9} \cdot 0.05}{10 \pi^2} \\
Q = \frac{128 \times 0.05 \times 10^{-9}}{10} \\
Q = \frac{6.4 \times 10^{-9}}{10} \\
Q = 6.4 \times 10^{-10} \, Q = 10 Ļ 2 128 Ļ 2 Ć 1 0 ā 9 ā
0.05 ā Q = 10 128 Ć 0.05 Ć 1 0 ā 9 ā Q = 10 6.4 Ć 1 0 ā 9 ā Q = 6.4 Ć 1 0 ā 10 C
Q = 64 Ć 10 ā 10 ā \\
Q = 64 \times 10^{-10} \, Q = 64 Ć 1 0 ā 10 C
Q = 64 Ć 10 ā 6 ā \\
Q = 64 \times 10^{-6} \, Q = 64 Ć 1 0 ā 6 C
Q = 64 ā Ī¼ \\
Q = 64 \, \mu Q = 64 Ī¼ C
Therefore, the total charge flowing through the coil is
C
This corresponds to Option 2.