Numerical Problems Based on Current Carrying Circular Loops for Class 12 Physics

Given parameters: Number of turns $N = 100$, Radius $R = 10\text{ cm} = 0.1\text{ m}$, Current $I = 1\text{ A}$.

The magnetic field at the centre of a circular coil is given by:

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

Using $\pi \approx 3.14$:

Given: $N = 1$, Current $I = 5.0\text{ A}$, Magnetic field $B = 0.20\text{ mT} = 0.20 \times 10^{-3}\text{ T}$.

Using the formula $B = \frac{\mu_0 I}{2R}$, rearrange to solve for the radius $R$:

Convert meters to centimeters:

To cancel the Earth’s magnetic field, the field produced by the coil at its centre must be equal and opposite to the horizontal component of the Earth’s field ($B_H$).

Given: $B = B_H = 1.86 \times 10^{-5}\text{ T}$, $N = 50$, $R = 2.54\text{ cm} = 2.54 \times 10^{-2}\text{ m}$.

Radius $R = 20\text{ cm} = 0.2\text{ m}$, Current $I = 10\text{ A}$.

The magnetic field at the centre of a full circular loop is $\frac{\mu_0 I}{2R}$. For a semicircular arc, the magnetic field is exactly half of that of a full loop:

Substitute the values:

Since the coils are identical and carry the same current $I$, the magnitude of the magnetic field produced by each coil at their common centre is equal:

The coils are in perpendicular planes, so their respective magnetic fields at the centre are also perpendicular to each other ($\vec{B_P} \perp \vec{B_Q}$).

The magnitude of the net magnetic field ($B_{net}$) is the vector sum of these two perpendicular fields:

The direction of the net magnetic field relative to the field of coil P is given by:

Thus, the net magnetic field acts at an angle of $45^\circ$ with respect to the magnetic field of either coil.