Ncert Exercises & Solutions

Find the current in the sliding rod AB (resistance=R) for the arrangement shown in the figure. B is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S is closed at time t=0.

Hint: The current is induced due to the change in magnetic flux.

Step 1: The conductor of length d moves with speed v, perpendicular to magnetic field B as shown in the figure. This produces motional emf across two ends of rod, which is given by=Bvd.

Since switch S is closed at time t=0, the capacitor is charged by this potential difference.

Step 2: Let Q(t) is a charge on the capacitor and current flows from A to B.

Now, the induced current;

I = \frac{Bvd}{R} - \frac{Q}{RC}

On rearranging the terms, we have;

\frac{Q}{RC} + \frac{dQ}{dt} = \frac{vBd}{R}

Step 3: This is the linear differential equation. On solving, we get;

\Rightarrow Q = vBdC + {Ae}^{- t / RC}

\Rightarrow Q = vBdC [1 - e^{- t / RC}] (At time t = 0, Q = 0 = A = - vBdC) .

Differentiating, we get,

I = \frac{vBd}{R} e^{- t / RC}

This is the required expression of current.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions