13.3 Solutions
627
= [M L
2 T
−2 Q
−1
]
∴
[emf] = [dφ B /dt]
13.78 The flux through the coil
φ = B · A = 15 × 10
−4
× 3 × 10
−12 sin(4 × 10
6 t)
ξ = −N
dφ
dt
= −200 × 15 × 10
−4
× 3 × 4 × 10
6
× 10
−12 cos(4 × 10
6 t)
= −3.6 × 10
−6 cos(4 × 10
6 t) V
13.79 ξ ∝
dφ
dt
or ∝
dB
dt
B = B 0 sin(ωt + φ)
∴
ξ ∝ ωB 0 cos(ωt + φ)
∴
ξ max (television)
ξ max (radio)
=
ω 1
ω 2
=
f 1
f 2
=
100
1
= 100
∵
B 0 is the same for both the waves.
13.80 The flux φ B enclosed by a loop of area A is given by φ B = B A, where B
is the magnetic field. Faraday’s law of induction says that the induced emf ξ
is a circuit equal to the negative rate at which the flux through the circuit is
changing. In symbols ξ = −dφ B /dt, the current being I = ξ/R, where R is
the resistance. The magnetic force on a straight wire is given by F = il × B:
φ B = B A = Blx (B⊥l)
By Faraday’s law
ξ = −
dϕ B
dt
= −
d
dt
(Blx) = −Bl
dx
dt
= Blv
i =
ξ
R
=
Blv
R
∴
Force F = ilB =
B 2 l 2 v
R
If the magnetic force acts as a resisting force then equation of motion will be
ma = −
B 2 l 2 v
R
or
m
dv
dt
= m
dv
ds
ds
dt
= mv
dv
ds
= −
B 2 l 2 v
R