32. When a conducting loop of resistan
from an external magnetic field act
ucting loop of resistance 10 12 and area 10 cm2 is remov
ating normally, the variation of induced
current-I in the loop with time t is as
de loop with time t is as shown in the figure.
Find the
(a) total charge passed through the loop
(b) change in magnetic flux through the top
(c) magnitude of the field applied.
(I in A)
0.5
(0,0)
2.0 (t in s)
17
5/5/2.
Answers
Answered by
2
Answer:
1 C
10 Weber
10^4 T
Explanation:
Answered by
0
(a) The total charge passed through the loop is 0.5 Columb.
(b) The change in magnetic flux through the top is 5 Weber.
(c) The magnitude of the field applied is 5000 Tesla.
Explanation:
The current is given as:
I = dq/dt ⇒ dq = I × dt
The total charge that passes through the area is:
q = 1 × (1/2 × 2 × 1/2)
∴ q = 0.5 C
The magnetic flux is given by the equation:
ΔQ = Δφ/R ⇒ Δφ = ΔQ × R
On substituting the values, we get,
Δφ = 1/2 × 10
∴ Δφ = 5 Wb
The magnetic field is given by the equation:
Δφ = B × ΔA ⇒ B = Δφ/ΔA
On substituting the values, we get,
B = 5/0.001
∴ B = 5000 T
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