CBSE Class 12 Physics Chapter 5: Magnetism and Matter NCERT Solutions

We are given the following information:

• Surface charge density of the earth, $\backslash sigma\; =\; 10^\{-9\}\; \backslash text\{\; C\; m\}^\{-2\}$.
• Current over the entire globe, $\backslash text\{I\}\; =\; 1800\; \backslash text\{\; A\}$.
• Radius of the earth, $r\; =\; 6.37\; \backslash times\; 10^6\; \backslash text\{\; m\}$.

First, we calculate the total surface area of the earth. The formula for the surface area of a sphere is $A\; =\; 4\backslash pi\; r^2$.

$$A\; =\; 4\backslash pi\; r^2\; =\; 4\backslash pi\; \backslash times\; (6.37\; \backslash times\; 10^6\; \backslash text\{\; m\})^2$$

$$A\; \backslash approx\; 4\; \backslash times\; 3.14159\; \backslash times\; (40.5769\; \backslash times\; 10^\{12\})\; \backslash text\{\; m\}^2$$

$$A\; \backslash approx\; 5.09\; \backslash times\; 10^\{14\}\; \backslash text\{\; m\}^2$$

Next, we find the total charge (q) on the earth's surface using the surface charge density and the calculated surface area:

$$q\; =\; \backslash sigma\; \backslash times\; A$$

$$q\; =\; (10^\{-9\}\; \backslash text\{\; C\; m\}^\{-2\})\; \backslash times\; (5.09\; \backslash times\; 10^\{14\}\; \backslash text\{\; m\}^2)$$

$$q\; =\; 5.09\; \backslash times\; 10^\{5\}\; \backslash text\{\; C\}$$

The current (I) is the rate of flow of charge, given by $I\; =\; \backslash frac\{q\}\{t\}$, where t is the time taken. To find the time required to neutralize the earth's surface, we rearrange the formula to $t\; =\; \backslash frac\{q\}\{I\}$.

$$t\; =\; \backslash frac\{5.09\; \backslash times\; 10^\{5\}\; \backslash text\{\; C\}\}\{1800\; \backslash text\{\; A\}\}$$

$$t\; \backslash approx\; 282.77\; \backslash text\{\; seconds\}$$

Therefore, it would take approximately 282.77 seconds to neutralize the earth's surface if there were no mechanism to sustain the electric field.

We are given:

• Number of secondary cells, $n\; =\; 6$.
• Emf of each cell, $E\; =\; 2.0\; \backslash text\{\; V\}$.
• Internal resistance of each cell, $r\; =\; 0.015\; \backslash Omega$.
• External resistance, $R\; =\; 8.5\; \backslash Omega$.

When cells are joined in series, their total emf is the sum of individual emfs, and their total internal resistance is the sum of individual internal resistances.

Total emf of the combination, $E\_\{total\}\; =\; n\; \backslash times\; E\; =\; 6\; \backslash times\; 2.0\; \backslash text\{\; V\}\; =\; 12.0\; \backslash text\{\; V\}$.

Total internal resistance, $r\_\{total\}\; =\; n\; \backslash times\; r\; =\; 6\; \backslash times\; 0.015\; \backslash Omega\; =\; 0.09\; \backslash Omega$.

The total resistance in the circuit is the sum of the external resistance and the total internal resistance: $R\_\{total\}\; =\; R\; +\; r\_\{total\}$.

$$R\_\{total\}\; =\; 8.5\; \backslash Omega\; +\; 0.09\; \backslash Omega\; =\; 8.59\; \backslash Omega$$

The current drawn from the supply (I) is given by Ohm's law for the entire circuit: $I\; =\; \backslash frac\{E\_\{total\}\}\{R\_\{total\}\}$.

$$I\; =\; \backslash frac\{12.0\; \backslash text\{\; V\}\}\{8.59\; \backslash Omega\}\; \backslash approx\; 1.397\; \backslash text\{\; A\}$$

Rounding to two decimal places, the current drawn is approximately $1.40\; \backslash text\{\; A\}$.

The terminal voltage (V) of the supply is the voltage across the external resistance R. It can be calculated as $V\; =\; I\; \backslash times\; R$.

$$V\; =\; 1.397\; \backslash text\{\; A\}\; \backslash times\; 8.5\; \backslash Omega\; \backslash approx\; 11.87\; \backslash text\{\; V\}$$

Alternatively, the terminal voltage can be calculated as the total emf minus the voltage drop across the internal resistance: $V\; =\; E\_\{total\}\; -\; I\; \backslash times\; r\_\{total\}$.

$$V\; =\; 12.0\; \backslash text\{\; V\}\; -\; (1.397\; \backslash text\{\; A\}\; \backslash times\; 0.09\; \backslash Omega)\; \backslash approx\; 12.0\; \backslash text\{\; V\}\; -\; 0.1257\; \backslash text\{\; V\}\; \backslash approx\; 11.87\; \backslash text\{\; V\}$$

Thus, the current drawn from the supply is approximately 1.40 A and the terminal voltage is approximately 11.87 V.

We are given:

• Emf of the secondary cell, $E\; =\; 1.9\; \backslash text\{\; V\}$.
• Internal resistance of the cell, $r\; =\; 380\; \backslash Omega$.

The maximum current that can be drawn from a cell occurs when the external resistance is zero (a short circuit). In this case, the current is given by Ohm's law applied to the cell itself: $I\_\{max\}\; =\; \backslash frac\{E\}\{r\}$.

$$I\_\{max\}\; =\; \backslash frac\{1.9\; \backslash text\{\; V\}\}\{380\; \backslash Omega\}$$

$$I\_\{max\}\; =\; 0.005\; \backslash text\{\; A\}$$

The maximum current that can be drawn from this cell is 0.005 A (or 5 mA).

A car's starting motor requires a very large current, typically in the range of hundreds of amperes, to crank the engine. Since the maximum current this cell can provide is only 0.005 A, it is far too small to drive the starting motor of a car.

Therefore, the cell cannot be used to start a car motor.