Respuesta :
Answer:
The induced current and the power dissipated through the resistor are 0.5 mA and [tex]7.5\times10^{-7}\ Watt[/tex].
Explanation:
Given that,
Distance = 1.0 m
Resistance = 3.0 Ω
Speed = 35 m/s
Angle = 53°
Magnetic field [tex]B=5.0\times10^{-5}\ T[/tex]
(a). We need to calculate the induced emf
Using formula of emf
[tex]E = Blv\sin\theta[/tex]
Where, B = magnetic field
l = length
v = velocity
Put the value into the formula
[tex]E=5.0\times10^{-5}\times1.0\times35\sin53^{\circ}[/tex]
[tex]E=1.398\times10^{-3}\ V[/tex]
We need to calculate the induced current
[tex]E =IR[/tex]
[tex]I=\dfrac{E}{R}[/tex]
Put the value into the formula
[tex]I=\dfrac{1.398\times10^{-3}}{3.0}[/tex]
[tex]I=0.5\ mA[/tex]
(b). We need to calculate the power dissipated through the resistor
Using formula of power
[tex]P=I^2 R[/tex]
Put the value into the formula
[tex]P=(0.5\times10^{-3})^2\times3.0[/tex]
[tex]P=7.5\times10^{-7}\ Watt[/tex]
Hence, The induced current and the power dissipated through the resistor are 0.5 mA and [tex]7.5\times10^{-7}\ Watt[/tex].
Answer:
a)
4.7 x 10⁻⁴ A
b)
6.63 x 10⁻⁷ Watt
Explanation:
L = distance between the two rails = 1.0 m
R = Resistance = 3.0 Ω
v = speed = 35 m/s
B = magnetic field = 5.0 x 10⁻⁵ T
θ = angle = 53°
Induced current is given as
[tex]i = \frac{BLvSin\theta }{R}[/tex]
[tex]i = \frac{(5\times 10^{-5})(1.0)(35)Sin53 }{3.0}[/tex]
i = 4.7 x 10⁻⁴ A
b)
Power dissipated is given as
P = i² R
P = (4.7 x 10⁻⁴)² (3)
P = 6.63 x 10⁻⁷ Watt
