CBSE Class 12 Physics Chapter 4: Moving Charges and Magnetism NCERT Solutions

We are given the following information for a circular coil:

• Number of turns, $n\; =\; 100$
• Radius of each turn, $r\; =\; 8.0\; \backslash text\{\; cm\}\; =\; 0.08\; \backslash text\{\; m\}$
• Current flowing in the coil, $I\; =\; 0.40\; \backslash text\{\; A\}$

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

Now, we substitute the given values into the formula:

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

First, let's simplify the numerator and denominator:

Numerator: $4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash times\; 100\; \backslash times\; 0.40\; =\; 160\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m\}$

Denominator: $2\; \backslash times\; 0.08\; \backslash text\{\; m\}\; =\; 0.16\; \backslash text\{\; m\}$

Now, divide the numerator by the denominator:

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

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

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

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

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

Rounding to two significant figures as per the given radius (8.0 cm):

$B\; \backslash approx\; 3.1\; \backslash times\; 10^\{-4\}\; \backslash text\{\; T\}$

Hence, the magnitude of the magnetic field at the centre of the coil is approximately $3.1\; \backslash times\; 10^\{-4\}$ 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, $r\; =\; 20\; \backslash text\{\; cm\}\; =\; 0.20\; \backslash text\{\; m\}$

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:

$$B\; =\; \backslash frac\{(4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m\; A\}^\{-1\})\; \backslash times\; 35\; \backslash text\{\; A\}\}\{2\backslash pi\; \backslash times\; 0.20\; \backslash text\{\; m\}\}$$

We can cancel out $2\backslash pi$ from the numerator and denominator:

$$B\; =\; \backslash frac\{2\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m\; A\}^\{-1\}\; \backslash times\; 35\; \backslash text\{\; A\}\}\{0.20\; \backslash text\{\; m\}\}$$

Now, perform the multiplication and division:

$B\; =\; \backslash frac\{70\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m\}\}\{0.20\; \backslash text\{\; m\}\}$

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

To express this in standard scientific notation:

$B\; =\; 3.5\; \backslash times\; 10^2\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\}$

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

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