Resistor network with mirror symmetry

JEE Advanced 2012 Paper 1, Question 12

For the resistance network shown in the figure, choose the correct option(s).

Rendered by QuickLaTeX.com

  1. The current through PQ is zero.
  2. I_1 = 3 A.
  3. The potential at S is less than that at Q.
  4. I_2 = 2 A.

Solution

We can speed up the solution by noticing the mirror symmetry of the circuit. Let the current I_1 split into two parts I_a and I_b at the node X (see figure). Similarly, let the currents going into node Y be I_a' and I_b'. If we were to reverse the polarity of the 12 \, {\rm V} battery connected to the circuit, all currents would simply reverse direction, but their magnitudes would remain the …

Continue Reading

Terminal velocity in a magnetic field

A copper connector of mass m slides down two smooth copper bars, set at an angle \alpha to the horizontal, due to gravity (see figure). At the top the bars are interconnected through a resistance R. The separation between the bars is \ell. The system is located in a uniform magnetic field of induction B, perpendicular to the plane in which the connector slides. The resistance of the bars, the connector and the sliding contacts, as well as the self-inductance of the loop are assumed to be negilible. If the connector is released from rest at t=0,

  1. Find the velocty v(t) of
Continue Reading

Electromagnetic induction in a twisted loop

JEE Advanced 2017 Paper 1, Question 5

A circular insulated copper wire loop is twisted to form two loops of area A and 2 A as shown in the figure. At the point of crossing the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform magnetic field {\bf B} points into the plane of the paper. At t=0, the loop starts rotating about the common diameter as axis with a constant angular velocity \omega in the magnetic field. Which of the following options is/are correct?

Rendered by QuickLaTeX.com

  1. The emf induced in the loop is proportional to the sum
Continue Reading

Coriolis effect and angular momentum

Imagine a mass m moving on the surface of a rotating sphere. For instance, the mass could be parcel of air moving away from a high pressure region in the Earth’s atmosphere. It experiences a Coriolis force which, in the example shown in the figure below, pushes it from its original trajectory (orange) to move eastward (blue). Why does this happen, and how do we understand it intuitively?

Rendered by QuickLaTeX.com

Formally, the Coriolis force on m is given by

(1)   \begin{equation*}   {\bf F}_{\rm Coriolis} = - 2 m {\bf \Omega} \times {\bf v}_{\rm rot} ,  \end{equation*}

where {\bf \Omega} is the angular velocity of the rotating frame (Earth), and {\bf v}_{\rm rot} is the velocity of m as seen by an observer on the Earth’s …

Continue Reading

Missing energy in a rope and a capacitor

Consider a uniform rope of mass density \lambda coiled on a smooth horizontal table. One end is pulled straight up with a constant speed v_0 as shown.

  1. Find the force exerted on the end of the rope as function of the height y.
  2. Compare the power delivered to the rope with the rate of change of the rope’s mechanical energy.

(This is a problem from chapter 5 of Kleppner and Kolenkow)

Rendered by QuickLaTeX.com

To find the force exerted at the top end, note that if we were to pull up a fixed mass with constant velocity v_0, the total force on the mass should …

Continue Reading