Two long, straight wires carry equal currents in opposite directions as shown in the figure. The separation between the wires is 5.0 cm$5.0\text{ cm}$. The magnitude of the magnetic field at a point P$P$ midway between the wires is ______ $\mu\text{T}$. (Given : μ₀ = 4π × 10⁻⁷ T ⁻¹$\mu_{0} = 4\pi \times 10^{-7}\text{ T}\cdot\text{m}\cdot\text{A}^{-1}$, and each wire carries a current of 10 A$10\text{ A}$).

Related Formula

The magnetic force on a moving charge is given by the Lorentz force equation:

Core Logic

Given data:

• Charge, q = 4.0 = 4.0 × 10⁻⁶ C$q = 4.0\ \mu\text{C} = 4.0 \times 10^{-6}\text{ C}$
• Velocity vector, v = (4.0 × 10⁶ j) ms⁻¹$\vec{v} = (4.0 \times 10^{6}\hat{j})\text{ ms}^{-1}$
• Magnetic field vector, B = 2 k T$\vec{B} = 2\hat{k}\text{ T}$

Step 1: Calculate the Force Vector

Substitute the values into the force equation:

Since j × k = i$\hat{j} \times \hat{k} = \hat{i}$:

Comparing this to x i N$x\hat{i}\text{ N}$, we find:

x = 32$x = 32$

Pattern Recognition

Cross-product check: j × k = i$\hat{j} \times \hat{k} = \hat{i}$. The product of the scalar terms (4.0 × 10⁻⁶) × (4.0 × 10⁶) × 2$(4.0 \times 10^{-6}) \times (4.0 \times 10^6) \times 2$ immediately simplifies to 4 × 4 × 2 = 32$4 \times 4 \times 2 = 32$.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Related Formula

Core Logic

For velocity to remain constant, the net force must be zero:

Let's evaluate the cases:

• Case (A): If E=0$E=0$ and B=0$B=0$, F = 0$\vec{F} = 0$. (Possible)
• Case (B): If E=0$E=0$ and B ≠ 0$B \neq 0$, the magnetic force is zero if v$\vec{v}$ is parallel or antiparallel to B$\vec{B}$ (i.e., v × B = 0$\vec{v} \times \vec{B} = 0$). (Possible)
• Case (C): If E ≠ 0$E \neq 0$ and B=0$B=0$, F = q E ≠ 0$\vec{F} = q\vec{E} \neq 0$, velocity must change. (Not possible)
• Case (D): If E ≠ 0$E \neq 0$ and B ≠ 0$B \neq 0$, the electric and magnetic forces can balance each other perfectly if q E = -q( v × B)$q\vec{E} = -q(\vec{v} \times \vec{B})$. (Possible)

Step 1: Conclusion

Hence, statements (A), (B), and (D) represent valid situations where velocity can remain constant.

Pattern Recognition

Velocity filter / velocity selector setups utilize crossed fields where electric fields perfectly balance magnetic components, an archetype of Case (D).

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Related Formula

To convert a galvanometer into an ammeter, a small resistance called a shunt (S$S$) is connected in parallel with the galvanometer (G$G$):

where, G$G$ = resistance of the galvanometer coil Ig$I_g$ = full-scale deflection current of the galvanometer I$I$ = total current to be measured S$S$ = shunt resistance

Core Logic

Given values:

Step 1: Calculate Shunt Resistance

Substituting these values into the shunt formula:

Therefore, the required value of the shunt resistance is 3.75 × 10⁻³ Ω$3.75 \times 10^{-3} \, \Omega$.

Pattern Recognition

Since I gg Ig$I \gg I_g$, the value I - Ig$I - I_g$ in the denominator can be approximated directly as I$I$ for a very accurate quick shortcut in choice selection: S ≈ (Ig G)/(I) = (0.03)/(8) = 3.75 × 10⁻³ Ω$S \approx \frac{I_g G}{I} = \frac{0.03}{8} = 3.75 \times 10^{-3} \, \Omega$.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Related Formula

For a shunted galvanometer, the potential difference across the galvanometer is equal to the potential difference across the shunt resistor:

Core Logic

Let the current sensitivity per division be x$x$.

• Initially, without the shunt, the total current I$I$ gives a full scale deflection of 25 divisions:

I = 25x$I = 25x$



• When the shunt (S = 24 Ω$S = 24 \, \Omega$) is connected in parallel, the current passing through the galvanometer branch (Ig$I_g$) corresponds to a deflection of 5 divisions:

Ig = 5x$I_g = 5x$

Step 1: Find Current through the Shunt

The current remaining for the shunt branch is:

Step 2: Equate Potential Drops

Applying the parallel branch condition:

Dividing both sides by 5x$5x$:

Therefore, the resistance of the galvanometer coil is 96 Ω$96 \, \Omega$.

Pattern Recognition

The deflection is directly proportional to the branch current. A drop from 25 divisions to 5 divisions means only (5)/(25) = (1)/(5)$\frac{5}{25} = \frac{1}{5}$ of the total current flows through the galvanometer, leaving (4)/(5)$\frac{4}{5}$ to go through the shunt. Hence, the galvanometer resistance must be 4 times the shunt resistance.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism