Question

A metal rod of resistance 15 ohm is moved to the right at a constant speed 60 cm/s along
two parallel conducting rails 25cm apart and shorted at one end. A magnetic field of
magnitude 0.35 T points into the page. Calculate the induced emf..

Answer 1

Answer:

(a) Equation E=

dt

dΦ

B

=

dt

d

BLx=BL

dt

dx

=BLv leads to

ε=BLv=(0.350T)(0.250m)(0.55m/s)=0.0481V

(b) By Ohm's law, the induced current is

i=0.0481V/18.0Ω=0.00267A

By Lenz's law, the current is clockwise in Figure.

(c) Equation P=i

2

R (resistive dissipation) leads to P=i

2

R=0.000129W

Answer 2

A metal rod of resistance 15Ω is moved to the right at a constant speed 60 cm/s along two parallel conducting rails 25cm apart and shorted at one end. A magnetic field of magnitude 0.35 T points into the page.

We have to find the induced emf.

magnitude of Induced emf is the the rate of change of magnetic flux.

$\implies\textit{Emf}=\frac{d\Phi}{dt}$

$=\frac{d(BLx)}{dt}$

$=BL\frac{dx}{dt}$

= BLv

here, B = 0.35 T , L = 25 cm = 0.25 m and v = 60 cm = 0.6 m/s

= 0.35 × 0.25 × 0.6 Volt

= 0.0525 volt

Therefore the induced emf would be 0.0525 volt.