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NCERT Class XII Chapter
Electromagnetic Induction
Questions With Solutions

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Question : 16
Total: 17
(a) Obtain an expression for mutual inductance between a long straight wire and a square loop of side ‘a’ as shown in figure.
(b) Now assume that straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m s–1.
Calculate the induced emf in the loop at the instant when x = 0.2 m.
Take a = 0.1m and assume that loop has a large resistance.

Solution:  

(a) As the magnetic field will be variable with distance from long straight wire, so the flux through square loop can be calculated by integration.
Let us assume a width ‘dr’ of the square loop at a distance ‘r’ from straight wire
B =
µ0
4π
2I
r
 or ϕ = B . Adr =
µ0
4π
2I
r
 adr
As ϕ = MI ⇒ MI =
µ0Ia
2π
loge (1 + a/x)
M =
µ0a
2π
loge (1 + a/x)
Total flux associated with square loop
ϕ = ∫ dϕ =
µ0
4π
2Ia‌
x+a
∫
x
dr
r
 or ϕ =
µ0
4π
 2Ia [loger]xx+a
ϕ =
µ0
4π
2Ia[loge
x+a
x
] or ϕ =
µ0Ia
2π
loge (1 + a/x)
(b) The square loop is moving right with a constant speed v, the instantaneous flux can be taken as
ϕ =
µ0Ia
2π
loge (1 + a/x)
Induced emf, ε = –
dϕ
dt
 = -
dϕ
dx
dx
dt
 = - v
dϕ
dx
ε = -
µ0Iav
2π
d(loge(1+
a
x
))
dx
 ⇒ ε = −
µ0Iav
2π
1
(1+
a
x
)
[−
a
x2
]
or ε =
µ0
2π
a2v
x(x+a)
 I or ε = 2 × 10−7
[0.1]2×10×50
0.2[0.2+0.1]
 = 1.67 × 10−5 V
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