Electromagnetic Induction

Power of lamp

or
decrease

Minimum possible resistance can be found in parallel arrangement and maximum possible resistance can be found in series arrangement.
Thus, n times


and n times

Let be the resistances of the bulbs. Power consumed

So, the power consumed in first bulb is

and power consumed in second bulb is

If these two bulbs are combined in series. So, the total resistance of the combination is

Hence, the power consumed in combination is

Now, from Eqs. (i) and (ii)



From E (iii)

First order derivation of thermo emf with respect to temperature
But we know that the thermoelectric power
So, the first order derivation of thermo emf produced in thermo-couple with respect to temperature gives thermoelectric power.

Given
and

and
We know that the heat produced

and heat produced per second



So, the ratio of heat produced per second in both the wires

On putting the values

Step 1: Understanding the Concept:
Self-inductance ( ) is a purely geometric property of a coil. It depends on the number of turns ( ), the cross-sectional area ( ), the length ( ), and the permeability of the core ( ).
Step 2: Detailed Explanation:
The formula for self-inductance of a solenoid is . While the magnetic flux ( ) depends on current ( ) according to , the coefficient itself does not change when the current changes. It is a constant for a given physical coil.
Step 3: Final Answer:
The self-inductance remains .

Step 1: Understanding the Concept:
The expression represents the magnetic energy density (energy per unit volume) stored in a magnetic field.
Step 2: Dimensional Analysis:
Energy Density =
Dimensions of Energy =
Dimensions of Volume =
$ $
Step 3: Final Answer:
The correct option is (a).

Step 1: Understanding the Concept:
Mutual inductance ( ) relates the flux linked with one coil due to the current in another: .
Step 2: Detailed Explanation:
Let current flow through the large outer coil. The magnetic field at the center is . Since , this field is approximately uniform over the area of the small coil. The flux linked with the small coil is : $ \Phi_1 = MI_2 M = \frac{\mu_0 \pi r_1^2}{2r_2} M \propto \frac{r_1^2}{r_2}$.
Step 3: Final Answer:
The correct option is (c).