Numerical Problems Based on Ampere’s Circuital Law for Class 12 Physics

Given parameters: Magnetic field $B = 20\text{ mT} = 20 \times 10^{-3}\text{ T}$.

Turn density ($n$) must be converted to turns per meter:

Using the formula for the magnetic field inside a long solenoid $B = \mu_0 n I$, solve for current ($I$):

Substitute the values ($\mu_0 = 4\pi \times 10^{-7}\text{ T m/A}$):

Wire radius $r = 0.5\text{ mm}$. Since the turns touch each other, the thickness of one turn equals the diameter of the wire.

The number of turns per unit length ($n$) is the reciprocal of the wire diameter:

Calculate the magnetic field ($B$) for a current $I = 5\text{ A}$:

Given parameters: Length $L = 50\text{ cm} = 0.5\text{ m}$, Magnetic field $B = 4.0 \times 10^{-2}\text{ T}$, Current $I = 8.0\text{ A}$.

Using the formula for magnetic field in terms of total turns $N$:

Substitute the given values (using $\pi = 3.14$):

Rounding to the nearest whole turn yields approximately 1990 turns.

Length $L = 1.0\text{ m}$, Diameter $= 3.0\text{ cm} \implies$ Radius $r = 1.5\text{ cm} = 0.015\text{ m}$. Current $I = 5.0\text{ A}$.

(i) Calculate magnetic field ($B$):
Total number of turns ($N$) is 5 layers of 850 turns each.

(ii) Calculate magnetic flux ($\Phi$):
The magnetic flux through the cross-sectional area ($A$) is given by $\Phi = B \cdot A$.

Rounded to two significant figures: $1.9 \times 10^{-5}\text{ Wb}$.

Length $L = 2.0\text{ m}$, Current $I = 5.0\text{ A}$.
Total number of turns $N = 5 \times 1000 = 5000\text{ turns}$.

Calculate the magnetic field ($B$):

Number of turns $N = 4200$, Current $I = 10\text{ A}$.
For the core of the toroid, the effective mean radius ($r$) is the average of the inner and outer radii:

(i) Inside the core of the toroid:

(ii) Outside the toroid:
According to Ampere’s Law, an Amperian loop outside the toroid encloses equal and opposite currents (current going in and coming out), making the net enclosed current zero. Thus, $B = 0$.

(iii) In the empty space surrounded by the toroid:
An Amperian loop inside the empty inner space encloses no current ($I_{enclosed} = 0$). Thus, $B = 0$.

Radius of conductor $R = 4\text{ cm} = 0.04\text{ m}$. Total current $I = 2\text{ A}$.
We need to find the field at distance $r = 3\text{ cm} = 0.03\text{ m}$. Since $r < R$, the point is inside the conductor.

Using Ampere’s Law for a uniform current distribution, the magnetic field inside a solid cylindrical conductor is proportional to the distance from the axis:

Substitute the given values: