Numerical Problems Based on Applications of Biot-Savart Law for Class 12 Physics

Given parameters: Current $I = 3\text{ A}$, Distance $r = 10\text{ cm} = 0.1\text{ m}$.

The magnetic field due to a long straight wire is given by:

Substituting the values (where $\frac{\mu_0}{2\pi} = 2 \times 10^{-7}\text{ T m/A}$):

Current $I = 12\text{ A}$, Magnetic field $B = 3 \times 10^{-5}\text{ T}$ (Note: $1\text{ Wb m}^{-2} = 1\text{ T}$).

Using the formula for magnetic field of a straight wire, $B = \frac{\mu_0 I}{2\pi r}$, and rearranging to solve for distance $r$:

The magnetic field $B$ produced by a long straight wire is inversely proportional to the distance $r$ from the wire ($B \propto \frac{1}{r}$).

Given: $r_1 = 4\text{ cm}$, $B_1 = 10^{-3}\text{ T}$, and $r_2 = 12\text{ cm}$.

We can set up a ratio since the current remains constant:

Given Magnetic flux density $B = 3 \times 10^{-5}\text{ T}$ and distance $r = 6\text{ cm} = 0.06\text{ m}$.

Using $B = \frac{\mu_0 I}{2\pi r}$, solve for current $I$:

At a neutral point, the magnetic field produced by the current-carrying wire exactly cancels out the horizontal component of the Earth’s magnetic field. Therefore, their magnitudes must be equal.

Given $r = 10\text{ cm} = 0.1\text{ m}$ and $B = B_H = 1.8 \times 10^{-4}\text{ T}$.

Given currents $I_1 = 10\text{ A}$ and $I_2 = 10\text{ A}$ in opposite directions. Total separation is $d = 5.0\text{ cm}$, so the distance from each wire to the midpoint P is $r = 2.5\text{ cm} = 0.025\text{ m}$.

Applying the Right-Hand Thumb Rule, the magnetic field at point P due to the first wire ($B_1$) and the second wire ($B_2$) both point in the same direction (perpendicular to the plane of the wires). Thus, the net magnetic field is their sum.

(Note: Some textbook editions may misprint the exponent in the answer key for this specific problem, but standard calculation confirms the magnitude of $10^{-4}$.)