Ncert Exercises & Solutions

A rod of mass m and resistance R slides smoothly over two parallel perfectly conducting wires kept sloping at an angle $θ$ with respect to the horizontal (figure). The circuit is closed through a perfect conductor at the top. There is a constant magnetic field B along the vertical direction. If the rod is initially at rest, find the velocity of the rod as a function of time.

Hint: The change in the magnetic flux results in the acceleration of the wire.

Step 1: Here, the component of the magnetic field perpendicular to the plane=

Bcosθ

Now, the conductor moves with speed v perpendicular to

Bcosθ

component of magnetic field. This causes motional emf across two ends of rod, which is given

ε

=v(

Bcosθ

This makes the flow of induced current,

i = \frac{v (Bcosθ) d}{R}

where, R is the resistance of rod. Now, current carrying rod experience force which is given by F=iBd (horizontally in backward direction).

Now, the component of magnetic force parallel to incline plane along upward direction=

F \cos θ = iBdcosθ = (\frac{v (Bcosθ) d}{R}) Bdcosθ where, v = \frac{dx}{dt}

Step 2: Also, the component of weight (mg) parallel to inclined plane along downward direction=mg sin

θ

Now, by Newton's second law of motion;

m \frac{d^{2} x}{{dt}^{2}} = mg sinθ - \frac{Bcosθd}{R} (\frac{dx}{dt}) \times (Bd) cosθ

\frac{dv}{dt} = gsinθ - \frac{B^{2} d^{2}}{mR} {(cosθ)}^{2} v

\frac{dv}{dt} + \frac{B^{2} d^{2}}{mR} {(cosθ)}^{2} v = gsinθ

Step 3: But, this is the linear differential equation.

On solving, we get;

v = \frac{g sinθ}{\frac{B^{2} d^{2} \cos^{2} θ}{mR}} + Aexp (- \frac{B^{2} d^{2}}{mR} (\cos^{2} θ) t)

A is a constant to be determined by initial conditions .

The require expression of velocity as a function of time is given by;

v = \frac{mgRsinθ}{B^{2} d^{2} \cos^{2} θ} (1 - \exp (- \frac{B^{2} d^{2}}{mR} (\cos^{2} θ) t))

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