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

We are given the following parameters for two concentric circular coils, X and Y:

• Radius of coil X, $r\_1\; =\; 16\; \backslash text\{\; cm\}\; =\; 0.16\; \backslash text\{\; m\}$
• Number of turns in coil X, $n\_1\; =\; 20$
• Current in coil X, $I\_1\; =\; 16\; \backslash text\{\; A\}$
• Radius of coil Y, $r\_2\; =\; 10\; \backslash text\{\; cm\}\; =\; 0.10\; \backslash text\{\; m\}$
• Number of turns in coil Y, $n\_2\; =\; 25$
• 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 centre 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, if we curl our fingers in the direction of the anticlockwise current, our thumb points towards the East. Therefore, the magnetic field $B\_1$ due to coil X is directed towards the East.

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

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

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

For coil Y, the current is clockwise when viewed from the west. Using the right-hand rule, if we curl our fingers in the direction of the clockwise current, our thumb points towards the West. Therefore, the magnetic field $B\_2$ due to coil Y is directed towards the West.

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

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

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

Since the magnetic fields $B\_1$ (East) and $B\_2$ (West) are in opposite directions, the net magnetic field $B$ at the centre is the difference between their magnitudes. Coil Y produces a stronger field ($B\_2$) than coil X ($B\_1$).

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 $B\_2$ is greater than $B\_1$, 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 $B\; =\; 100\; \backslash text\{\; G\}\; =\; 100\; \backslash times\; 10^\{-4\}\; \backslash text\{\; T\}$ in a specific region.

The magnetic field inside a long solenoid is given by the formula $B\; =\; \backslash mu\_0\; n\; I$, where:

• $B$ is the magnetic field strength.
• $\backslash mu\_0\; =\; 4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m\; A\}^\{-1\}$ is the permeability of free space.
• $n$ is the number of turns per unit length.
• $I$ is the current flowing through the solenoid.

We are given the following constraints:

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

We need to find values for $n$ and $I$ such that $B\; =\; \backslash mu\_0\; n\; I$ and $n\; \backslash le\; n\_\{max\}$, $I\; \backslash le\; I\_\{max\}$.

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

$$I\; =\; \backslash frac\{B\}\{\backslash mu\_0\; n\}\; =\; \backslash frac\{100\; \backslash times\; 10^\{-4\}\; \backslash text\{\; T\}\}\{(4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m\; A\}^\{-1\})\; \backslash times\; (1000\; \backslash text\{\; turns\; m\}^\{-1\})\}$$

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

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

$$I\; \backslash approx\; \backslash frac\{25\}\{3.14159\}\; \backslash text\{\; A\}\; \backslash approx\; 7.96\; \backslash text\{\; A\}$$

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, one possible design for the solenoid is:

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

Alternatively, we could choose a different combination of $n$ and $I$ that satisfies the constraints. For instance, if we choose a current $I\; =\; 10\; \backslash text\{\; A\}$ (which is less than 15 A), the required number of turns per unit length would be:

$$n\; =\; \backslash frac\{B\}\{\backslash mu\_0\; I\}\; =\; \backslash frac\{100\; \backslash times\; 10^\{-4\}\; \backslash text\{\; T\}\}\{(4\backslash pi\; \backslash times\; 10^\{-7\}\; \backslash text\{\; T\; m\; A\}^\{-1\})\; \backslash times\; 10\; \backslash text\{\; A\}\}\; =\; \backslash frac\{100\; \backslash times\; 10^\{-4\}\}\{40\backslash pi\; \backslash times\; 10^\{-7\}\}\; \backslash text\{\; m\}^\{-1\}$$

$$n\; =\; \backslash frac\{100\; \backslash times\; 10^\{-4\}\}\{4\backslash pi\; \backslash times\; 10^\{-6\}\}\; \backslash text\{\; m\}^\{-1\}\; =\; \backslash frac\{100\}\{4\backslash pi\}\; \backslash times\; 10^2\; \backslash text\{\; m\}^\{-1\}\; =\; \backslash frac\{25\}\{\backslash pi\}\; \backslash times\; 100\; \backslash text\{\; m\}^\{-1\}\; \backslash approx\; 796\; \backslash text\{\; m\}^\{-1\}$$

Since $n\; \backslash approx\; 796\; \backslash text\{\; m\}^\{-1\}$ is less than $n\_\{max\}\; =\; 1000\; \backslash text\{\; turns\; m\}^\{-1\}$, this is also a valid design.

Suggested Design Particulars:

Option 1: Use the maximum number of turns per unit length.

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

Option 2: Use a current less than the maximum capacity.

• Current, $I\; =\; 10\; \backslash text\{\; A\}$
• Number of turns per unit length, $n\; \backslash approx\; 796\; \backslash text\{\; turns\; m\}^\{-1\}$

Both options satisfy the given constraints and will produce the required magnetic field of 100 G.