CBSE Class 12 Physics Chapter 8: Moving Charges and Magnetism NCERT Solutions

We are given the following information for a circular coil of wire:

• Number of turns, $n\; =\; 100$
• Radius of each turn, $r\; =\; 8.0\; \backslash text\{\; cm\}$
• Current flowing in the coil, $I\; =\; 0.40\; \backslash text\{\; A\}$

First, we need to convert the radius from centimeters to meters:

$r\; =\; 8.0\; \backslash text\{\; cm\}\; =\; 8.0\; \backslash times\; 10^\{-2\}\; \backslash text\{\; m\}\; =\; 0.08\; \backslash text\{\; m\}$

The magnitude of the magnetic field (B) at the centre of a circular coil is given by the formula:

$$B\; =\; \backslash frac\{\backslash mu\_0\; n\; I\}\{2r\}$$

Where $\backslash mu\_0$ is the permeability of free space, and its value is $\backslash mu\_0\; =\; 4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m\; A\}^\{-1\}$.

Now, we substitute the given values into the formula:

$$B\; =\; \backslash frac\{(4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m\; A\}^\{-1\})\; \backslash times\; 100\; \backslash times\; (0.40\; \backslash text\{\; A\})\}\{2\; \backslash times\; (0.08\; \backslash text\{\; m\})\}$$

Let's simplify the expression:

$$B\; =\; \backslash frac\{4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash times\; 40\}\{0.16\}\; \backslash text\{\; T\}$$

$$B\; =\; \backslash frac\{160\backslash pi\; \backslash times\; 10^\{-7\}\}\{0.16\}\; \backslash text\{\; T\}$$

$$B\; =\; 1000\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\}$$

$$B\; =\; \backslash pi\; \backslash times\; 10^\{-4\}\; \backslash text\{\; T\}$$

Using the approximate value of $\backslash pi\; \backslash approx\; 3.14159$:

$$B\; \backslash approx\; 3.14159\; \backslash times\; 10^\{-4\}\; \backslash text\{\; T\}$$

Therefore, the magnitude of the magnetic field at the centre of the coil is approximately $3.14\; \backslash times\; 10^\{-4\}\; \backslash text\{\; T\}$.

We are given the following information for a long straight wire:

• Current in the wire, $I\; =\; 35\; \backslash text\{\; A\}$
• Distance of the point from the wire, $r\; =\; 20\; \backslash text\{\; cm\}$

First, convert the distance from centimeters to meters:

$r\; =\; 20\; \backslash text\{\; cm\}\; =\; 20\; \backslash times\; 10^\{-2\}\; \backslash text\{\; m\}\; =\; 0.2\; \backslash text\{\; m\}$

The magnitude of the magnetic field (B) at a perpendicular distance r from a long straight wire carrying current I is given by the formula:

$$B\; =\; \backslash frac\{\backslash mu\_0\; I\}\{2\backslash pi\; r\}$$

Where $\backslash mu\_0$ is the permeability of free space, with a value of $\backslash mu\_0\; =\; 4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m\; A\}^\{-1\}$.

Substitute the given values into the formula:

$$B\; =\; \backslash frac\{(4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m\; A\}^\{-1\})\; \backslash times\; (35\; \backslash text\{\; A\})\}\{2\backslash pi\; \backslash times\; (0.2\; \backslash text\{\; m\})\}$$

Simplify the expression:

$$B\; =\; \backslash frac\{2\; \backslash times\; 10^\{-7\}\; \backslash times\; 35\}\{0.2\}\; \backslash text\{\; T\}$$

$$B\; =\; \backslash frac\{70\; \backslash times\; 10^\{-7\}\}\{0.2\}\; \backslash text\{\; T\}$$

$$B\; =\; 350\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\}$$

$$B\; =\; 3.5\; \backslash times\; 10^\{-5\}\; \backslash text\{\; T\}$$

Hence, the magnitude of the magnetic field at a point 20 cm from the wire is $3.5\; \backslash times\; 10^\{-5\}\; \backslash text\{\; T\}$.