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

Answer:

The induced current and the power dissipated through the resistor are 0.5 mA and $7.5\times10^{-7}\ Watt$.

Explanation:

Given that,

Distance = 1.0 m

Resistance = 3.0 Ω

Speed = 35 m/s

Angle = 53°

Magnetic field $B=5.0\times10^{-5}\ T$

(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

$E=5.0\times10^{-5}\times1.0\times35\sin53^{\circ}$

$E=1.398\times10^{-3}\ V$

We need to calculate the induced current

$E =IR$

$I=\dfrac{E}{R}$

Put the value into the formula

$I=\dfrac{1.398\times10^{-3}}{3.0}$

$I=0.5\ mA$

(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

$P=(0.5\times10^{-3})^2\times3.0$

$P=7.5\times10^{-7}\ Watt$

Hence, The induced current and the power dissipated through the resistor are 0.5 mA and $7.5\times10^{-7}\ Watt$.

Answer 2

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

$i = \frac{BLvSin\theta }{R}$

$i = \frac{(5\times 10^{-5})(1.0)(35)Sin53 }{3.0}$

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