A current carrying is placed vertically and a particle of mass m with charge Q is released from rest. The particle moves along the axis of solenoid. If g is acceleration due to gravity then the acceleration (a) of the charged particle will satisfy :

Related Formula

For a semi-infinite wire segment at distance r$r$:

For a full circular loop at its center:

Core Logic

The total magnetic field at O is the vector sum of fields from three segments:

• The incoming semi-infinite wire (AB)
• The outgoing semi-infinite wire (DE)
• The nearly full circular loop (BCD)

Note: Based on the diagram, the loop is not fully closed, but geometrically it acts as a full circle subtracted by the gap. Typically this standard shape treats the circular part as a full circle and the straight wires as two semi-infinite wires.

Step 1: Adding the Vector Components

Applying the Right Hand Rule:

• Segment AB: current flows along +x$+x$, position vector to O is +y$+y$. d l × r = i × j = k$d\vec{l} \times \vec{r} = \hat{i} \times \hat{j} = \hat{k}$. Wait, the diagram shows the loop in the x-y$x-y$ plane. Let's re-examine axes. Based on standard convention, if current is in xy$xy$ plane, field is in z$z$ (k$\hat{k}$) direction. The solution shows vectors in i$\hat{i}$. This means the axes are drawn such that the loop is in the y-z$y-z$ plane. Yes, the provided axes show x$x$ pointing out, y$y$ to the right, z$z$ upwards.
• Segment AB (current along y$y$ axis): B$B$ at origin is along +x$+x$ (i$\hat{i}$).
• Segment DE (current along y$y$ axis): B$B$ at origin is along +x$+x$ (i$\hat{i}$).
• Circular Loop (current clockwise in y-z$y-z$ plane): B$B$ at origin points inwards, i.e., -x$-x$ (- i$-\hat{i}$).

Step 2: Final Conclusion

Pattern Recognition

Always separate complex wire geometries into standard segments: infinite wires, semi-infinite wires, and arcs. Use the Right-Hand Rule carefully with the given explicit coordinate frame to avoid sign errors.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Related Formula

Core Logic

For an equilateral triangle, the perpendicular distance d$d$ from centroid to any side is given by d = a2√(3)$d = \frac{a}{2\sqrt{3}}$, where a = 4√(3) cm$a = 4\sqrt{3} \, \mathrm{cm}$.

The angles subtended by the side at the centroid are θ₁ = 60°$\theta_1 = 60^{\circ}$ and θ₂ = 60°$\theta_2 = 60^{\circ}$.

Step 1: Field due to one side

Step 2: Total Magnetic Field

Since there are 3 identical sides and their field vectors point in the same direction at the centroid:

Comparing with α × 10⁻⁵T$\alpha \times 10^{-5}\mathrm{T}$, we get α = 3√(3)$\alpha = 3\sqrt{3}$.

Pattern Recognition

For regular polygons of n$n$ sides, Bcentroid = n · Bside$B_{\text{centroid}} = n \cdot B_{\text{side}}$. An equilateral triangle has n=3$n=3$, d = a/(2√(3))$d = a/(2\sqrt{3})$, and angles are always 60°$60^{\circ}$.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Related Formula

F = qvB$F = qvB$

Core Logic

For Magnetic induction (B):

So, A matches III.

For Magnetic Flux (φ$\phi$):

So, B matches IV.

For Magnetic Permeability (μ$\mu$):

So, C matches I.

For Self inductance (L): Using U = (1)/(2) LI²$U = \frac{1}{2} LI^{2}$,

Wait, the given option II is [ML² T⁻² A⁻²]$[ML^2 T^{-2} A^{-2}]$. Let's re-verify: Energy U = [ML² T⁻²]$U = [ML^2 T^{-2}]$. I² = [A²]$I^2 = [A^2]$. So L = [ML² T⁻² A⁻²]$L = [ML^2 T^{-2} A^{-2}]$. So, D matches II.

Step 1: Final Conclusion

A-III, B-IV, C-I, D-II. Option (4) is correct.

Pattern Recognition

Dimensional analysis of electromagnetic quantities frequently hinges on knowing formulas for force, flux, and energy. Deriving from F=qvB$F=qvB$ and U=(1)/(2)LI²$U=\frac{1}{2}LI^2$ is the fastest approach.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism Class 12 Physics: Electromagnetic Induction

Related Formula

Core Logic

The net magnetic field at O is the vector sum of fields from both loops. Since the currents are in the same relative orientation (clockwise from O), their magnetic fields at O will point in opposite directions.

Step 1: Superposition of Fields

Magnetic field due to loop Q (carrying 4I) towards Q, and due to loop P (carrying I) towards P.

Step 2: Final Calculation

The direction is towards Q because the field from the loop carrying 4I$4I$ is dominant.

Pattern Recognition

When symmetrical coils carry opposing fields along their axis at equidistance, you simply subtract their current multipliers (4I - I = 3I) and apply the standard axial magnetic field formula once.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism