Question

An experimental apparatus has two parallel horizontal metal rails separated by 1.0 m. A 3.0 Ω resistor is connected from the left end of one rail to the left end of the other. A metal axle with metal wheels is pulled toward the right along the rails at a speed of 35 m/s . Earth's uniform 5.0 × 10−5-T field points down at an angle of 53∘ below the horizontal.

(a) Determine the induced current.
(b) Determine the power dissipated through the resistor.

Answer 1

a)

4.7 x 10⁻⁴ A

b)

6.63 x 10⁻⁷ Watt

Step-by-step 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

`i = (BLvSin\theta )/(R)`

`i = ((5* 10^(-5))(1.0)(35)Sin53 )/(3.0)`

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

Answer 2

The induced current and the power dissipated through the resistor are 0.5 mA and
`7.5*10^(-7)\ Watt`.

Step-by-step explanation:

Given that,

Distance = 1.0 m

Resistance = 3.0 Ω

Speed = 35 m/s

Angle = 53°

Magnetic field
`B=5.0*10^(-5)\ T`

(a). We need to calculate the induced emf

Using formula of emf

`E = Blv\sin\theta`

Where, B = magnetic field

l = length

v = velocity

Put the value into the formula

`E=5.0*10^(-5)*1.0*35\sin53^(\circ)`

`E=1.398*10^(-3)\ V`

We need to calculate the induced current

`E =IR`

`I=(E)/(R)`

Put the value into the formula

`I=(1.398*10^(-3))/(3.0)`

`I=0.5\ mA`

(b). We need to calculate the power dissipated through the resistor

Using formula of power

`P=I^2 R`

Put the value into the formula

`P=(0.5*10^(-3))^2*3.0`

`P=7.5*10^(-7)\ Watt`

Hence, The induced current and the power dissipated through the resistor are 0.5 mA and
`7.5*10^(-7)\ Watt`.