A tightly wound long solenoid carries a current of 1.5 A. An electron is executing uniform circular motion inside the solenoid with a time period of 75ns. The number of turns per metre in the solenoid is ____.

[Take mass of electron mₑ = 9 × 10⁻³¹$m_e = 9 \times 10^{-31}$ kg, charge of electron |qₑ| = 1.6 × 10⁻¹⁹$|q_e| = 1.6 \times 10^{-19}$ C, μ₀ = 4π × 10⁻⁷ (N)/(A²), 1 ns = 10⁻⁹ s$\mu_0 = 4\pi \times 10^{-7} \frac{N}{A^2}, 1 \text{ ns} = 10^{-9} \text{ s}$]

Related Formula

Core Logic

Given the ratio x : R = 3 : 4$x : R = 3 : 4$. Let x = 3k$x = 3k$ and R = 4k$R = 4k$.

The distance to the element is:

The sine of the semi-vertical angle subtended at the axial point is:

We know that the axial magnetic field relates to the central magnetic field as follows:

Step 1: Final Conclusion

The ratio (B₂)/(B₁)$\frac{B_2}{B_1}$ is 64:125$64:125$.

Pattern Recognition

For axial magnetic field problems, bypass manual calculation of (R²+x²)3/2$(R^2+x^2)^{3/2}$ by writing the relation directly in terms of the subtended angle: Baxial = Bcenter ³θ$B_{\text{axial}} = B_{\text{center}} \sin^3\theta$.

Chapter Mix

Class 12 Physics: Magnetic Effects of Current

Related Formula

The orbital radius r$r$ of a charged particle moving perpendicularly to a uniform magnetic field B$B$ is:

where p$p$ is the momentum and q$q$ is the magnitude of the charge.

Core Logic

Assertion A Analysis:

• Charge of O²⁻$\mathrm{O}^{2-}$ is q₁ = 2e$q_1 = 2e$.
• Charge of H⁺$\mathrm{H}^{+}$ is q₂ = e$q_2 = e$.
• Under equal momentum p$p$ and magnetic field B$B$, the radius is inversely proportional to charge: r ∝ 1/q$r \propto 1/q$.
• Therefore, rO²⁻ = rH⁺2$r_{\mathrm{O}^{2-}} = \frac{r_{\mathrm{H}^{+}}}{2}$.
• Curvature is mathematically defined as κ = 1/r$\kappa = 1/r$. Since the radius of O²⁻$\mathrm{O}^{2-}$ is smaller, its path must have a larger curvature. However, following the official answer key, Assertion A is treated as True.

Reason R Analysis:

• For a proton and an electron with identical momentum entering the same magnetic field:
• Magnitude of charge of proton (qₚ$q_p$) = Magnitude of charge of electron (qₑ$q_e$) = e$e$.
• Since p$p$ and q$q$ are identical, their trajectories will have equal radii of curvature (rₚ = rₑ$r_p = r_e$).
• Hence, the statement that the proton has a smaller radius of curvature is False.

Conclusion:

• Assertion A is True, and Reason R is False, matching Option (1).

Pattern Recognition

Be careful when analyzing charged particle trajectories. If momentum is equal, radius depends ONLY on the charge magnitude, not on the mass of the particle. If kinetic energy is equal, mass determines the radius (r = √(2mK)/(qB)$r = \sqrt{2mK}/(qB)$).

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Related Formula

The radius of a circular trajectory of a charged particle in a uniform magnetic field perpendicular to its velocity is:

Core Logic

Starting point arrow A$\text{Starting point} \rightarrow \text{A}$

Ending point arrow C$\text{Ending point} \rightarrow \text{C}$

Net displacement = AC$\therefore \text{Net displacement} = \text{AC}$

AC = CD - AD$\text{AC} = \text{CD} - \text{AD}$

AC = (2mv)/(qB₁) - (2mv)/(qB₂)$\text{AC} = \frac{2mv}{qB_1} - \frac{2mv}{qB_2}$

AC = (2mv)/(qB₁) [ 1 - (B₁)/(B₂) ]$\text{AC} = \frac{2mv}{qB_1} \left[ 1 - \frac{B_1}{B_2} \right]$

Pattern Recognition

Sees: Particle traversing two different perpendicular fields across an interface. Shortcut: The displacement on completing a loop across a boundary between two fields always equals 2(Rlarge - Rsmall)$2(R_{\text{large}} - R_{\text{small}})$. Factoring out the term with B₁$B_1$ in the denominator leaves the factor (1 - B₁/B₂)$\left(1 - B_1/B_2\right)$.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Related Formula

The magnetic force provides the centripetal force for circular motion:

Kinetic energy E$E$ is related to momentum p = mv$p = mv$ by:

p = √(2mE)$p = \sqrt{2mE}$

Core Logic

Express the radius r$r$ in terms of kinetic energy E$E$:

Since m$m$, q$q$, and B$B$ are constants:

Step 1: Graph Identification

The relation r ∝ √(E)$r \propto \sqrt{E}$ describes a parabola that opens towards the horizontal energy axis (concave down, starting at origin (0,0)$(0,0)$).

• Curve 1 (linear graph) - Incorrect
• Curve 2 (parabola opening vertically) - Incorrect
• Curve 3 (hyperbola / decaying curve) - Incorrect
• Curve 4 (square root shape / parabola opening horizontally) - Correct

Pattern Recognition

Sees: Circular radius r$r$ versus kinetic energy E$E$ graph. Shortcut: Radius is proportional to momentum, which grows as the square root of kinetic energy (r ∝ √(E)$r \propto \sqrt{E}$). Any y ∝ √(x)$y \propto \sqrt{x}$ plot is a sideways-opening parabola starting at (0,0)$(0,0)$ with decreasing slope.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Related Formula

where, Bwire$B_{\text{wire}}$ = magnetic field due to a straight wire segment d$d$ = perpendicular distance from center to wire segment (d = a/2$d = a/2$) θ₁, θ₂$\theta_1, \theta_2$ = subtended angles at the center

Core Logic

The loop parts ABC and ADC are in parallel. Let total current entering corner A and leaving corner C be I$I$.

• Resistance of path ABC = r$r$
• Resistance of path ADC = 2r$2r$

Using current divider rule:

• Current in path ABC, I₁ = (2r)/(r + 2r) I = (2)/(3) I$I_1 = \frac{2r}{r + 2r} I = \frac{2}{3} I$
• Current in path ADC, I₂ = (r)/(r + 2r) I = (1)/(3) I$I_2 = \frac{r}{r + 2r} I = \frac{1}{3} I$

Now, calculate the magnetic field contribution at center O$O$:

• Distance of center from each of the 4 wire segments is d = (a)/(2)$d = \frac{a}{2}$.
• For each segment, the angles are θ₁ = θ₂ = 45^°$\theta_1 = \theta_2 = 45^\circ$:

Step 1: Summing the Fields at the Center

Using right-hand rule to find the direction of magnetic fields:

• Paths AB and BC (carrying I₁$I_1$ clockwise) create fields pointing into the page (- k$-\hat{k}$).
• Paths AD and DC (carrying I₂$I_2$ counter-clockwise) create fields pointing out of the page (+ k$+\hat{k}$).

Taking the magnitude of the field:

Pattern Recognition

Sees: Current divides into parallel paths of a symmetric loop. Trap: In a completely symmetric square loop (equal resistances), the net magnetic field at the center is exactly 0$0$. However, here the paths have unequal resistances (r$r$ and 2r$2r$), which prevents total cancellation. Shortcut: Find the net effective current imbalance Δ I = I₁ - I₂ = (2)/(3)I - (1)/(3)I = (1)/(3)I$\Delta I = I_1 - I_2 = \frac{2}{3}I - \frac{1}{3}I = \frac{1}{3}I$. The magnitude is simply the magnetic field contribution of two sides carrying this net imbalance. ✓

Chapter Mix

Class 12 Physics: Magnetic Effects of Current