CBSE Class 12 Physics Chapter 7: Moving Charges and Magnetism NCERT Solutions

We are given the following information:

• Radius of coil X, $r\_1\; =\; 16\; \backslash text\{\; cm\}\; =\; 0.16\; \backslash text\{\; m\}$
• Radius of coil Y, $r\_2\; =\; 10\; \backslash text\{\; cm\}\; =\; 0.10\; \backslash text\{\; m\}$
• Number of turns in coil X, $N\_1\; =\; 20$
• Number of turns in coil Y, $N\_2\; =\; 25$
• Current in coil X, $I\_1\; =\; 16\; \backslash text\{\; A\}$
• Current in coil Y, $I\_2\; =\; 18\; \backslash text\{\; A\}$
• Permeability of free space, $\backslash mu\_0\; =\; 4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m\; A\}^\{-1\}$

The magnetic field at the center of a circular coil is given by the formula $B\; =\; \backslash frac\{\backslash mu\_0\; N\; I\}\{2r\}$.

For coil X, the current is anticlockwise when viewed from the west. Using the right-hand rule, this current produces a magnetic field directed towards the East at the center.

The magnitude of the magnetic field due to coil X is:

$$B\_1\; =\; \backslash frac\{\backslash mu\_0\; N\_1\; I\_1\}\{2r\_1\}\; =\; \backslash frac\{(4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m\; A\}^\{-1\})\; \backslash times\; 20\; \backslash times\; 16\; \backslash text\{\; A\}\}\{2\; \backslash times\; 0.16\; \backslash text\{\; m\}\}$$

$$B\_1\; =\; \backslash frac\{4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash times\; 320\}\{0.32\}\; =\; 4\backslash pi\; \backslash times\; 10^\{-4\}\; \backslash text\{\; T\; (Towards\; East)\}$$

For coil Y, the current is clockwise when viewed from the west. Using the right-hand rule, this current produces a magnetic field directed towards the West at the center.

The magnitude of the magnetic field due to coil Y is:

$$B\_2\; =\; \backslash frac\{\backslash mu\_0\; N\_2\; I\_2\}\{2r\_2\}\; =\; \backslash frac\{(4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m\; A\}^\{-1\})\; \backslash times\; 25\; \backslash times\; 18\; \backslash text\{\; A\}\}\{2\; \backslash times\; 0.10\; \backslash text\{\; m\}\}$$

$$B\_2\; =\; \backslash frac\{4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash times\; 450\}\{0.20\}\; =\; 9\backslash pi\; \backslash times\; 10^\{-4\}\; \backslash text\{\; T\; (Towards\; West)\}$$

Since the magnetic fields $B\_1$ and $B\_2$ are in opposite directions (East and West), the net magnetic field at the center is the difference between their magnitudes. The direction of the net field will be the same as the direction of the larger field.

Net magnetic field, $B\; =\; |B\_2\; -\; B\_1|$

$$B\; =\; |9\backslash pi\; \backslash times\; 10^\{-4\}\; \backslash text\{\; T\}\; -\; 4\backslash pi\; \backslash times\; 10^\{-4\}\; \backslash text\{\; T\}|\; =\; 5\backslash pi\; \backslash times\; 10^\{-4\}\; \backslash text\{\; T\}$$

Substituting the value of $\backslash pi\; \backslash approx\; 3.14$:

$$B\; =\; 5\; \backslash times\; 3.14\; \backslash times\; 10^\{-4\}\; \backslash text\{\; T\}\; =\; 15.7\; \backslash times\; 10^\{-4\}\; \backslash text\{\; T\}\; =\; 1.57\; \backslash times\; 10^\{-3\}\; \backslash text\{\; T\}$$

Since $B\_2$ (towards West) is greater than $B\_1$ (towards East), the net magnetic field is directed towards the West.

Answer: The magnitude of the net magnetic field at the centre is $1.57\; \backslash times\; 10^\{-3\}\; \backslash text\{\; T\}$ and its direction is towards the West.

We need to design a solenoid that produces a uniform magnetic field of 100 G in a specific region. The given parameters are:

• Required magnetic field strength, $B\; =\; 100\; \backslash text\{\; G\}\; =\; 100\; \backslash times\; 10^\{-4\}\; \backslash text\{\; T\}\; =\; 10^\{-2\}\; \backslash text\{\; T\}$
• Maximum number of turns per unit length, $n\_\{max\}\; =\; 1000\; \backslash text\{\; turns\; m\}^\{-1\}$
• Maximum current carrying capacity, $I\_\{max\}\; =\; 15\; \backslash text\{\; A\}$
• Permeability of free space, $\backslash mu\_0\; =\; 4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m\; A\}^\{-1\}$

The magnetic field inside a long solenoid is given by the formula $B\; =\; \backslash mu\_0\; n\; I$. We need to find values of $n$ and $I$ such that $B\; =\; 10^\{-2\}\; \backslash text\{\; T\}$, and $n\; \backslash le\; n\_\{max\}$ and $I\; \backslash le\; I\_\{max\}$.

Let's consider the maximum possible turns per unit length, $n\; =\; 1000\; \backslash text\{\; turns\; m\}^\{-1\}$. We can then calculate the required current $I$ using the formula:

$$I\; =\; \backslash frac\{B\}\{\backslash mu\_0\; n\}\; =\; \backslash frac\{10^\{-2\}\; \backslash text\{\; T\}\}\{(4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m\; A\}^\{-1\})\; \backslash times\; (1000\; \backslash text\{\; turns\; m\}^\{-1\})\}$$

$$I\; =\; \backslash frac\{10^\{-2\}\}\{4\backslash pi\; \backslash times\; 10^\{-4\}\}\; \backslash text\{\; A\}\; =\; \backslash frac\{100\}\{4\backslash pi\}\; \backslash text\{\; A\}\; =\; \backslash frac\{25\}\{\backslash pi\}\; \backslash text\{\; A\}$$

Calculating the value of $\backslash frac\{25\}\{\backslash pi\}$:

$$I\; \backslash approx\; \backslash frac\{25\}\{3.14\}\; \backslash text\{\; A\}\; \backslash approx\; 7.96\; \backslash text\{\; A\}$$

Since the required current $I\; \backslash approx\; 7.96\; \backslash text\{\; A\}$ is less than the maximum allowed current $I\_\{max\}\; =\; 15\; \backslash text\{\; A\}$, this design is feasible.

Therefore, appropriate design particulars for the solenoid are:

• Number of turns per unit length, $n\; =\; 1000\; \backslash text\{\; turns\; m\}^\{-1\}$
• Current, $I\; \backslash approx\; 7.96\; \backslash text\{\; A\}$

The solenoid should be wound to have 1000 turns per meter and operated with a current of approximately 7.96 A to produce a magnetic field of 100 G. The length of the solenoid should be chosen such that it provides a uniform field over the required region of 10 cm linear dimension. For a long solenoid, the field is uniform in the central region.