A loop ABCDA, carrying current I = 12~A$I = 12\mathrm{~A}$, is placed in a plane, consists of two semi-circular segments of radius R₁ = 6π~m$R_1 = 6\pi\mathrm{~m}$ and R₂ = 4π~m$R_2 = 4\pi\mathrm{~m}$. The magnitude of the resultant magnetic field at center O is k× 10⁻⁷~T$k\times 10^{-7}\mathrm{~T}$ The value of k is ________. (Given μ₀=4π×10⁻⁷~T· m· A⁻¹$\mu_{0}=4\pi\times10^{-7}\mathrm{~T\cdot m\cdot A}^{-1}$)
A current loop containing two concentric semicircular segments of radii R1 and R2 connected by radial straight segments, with common center O.
Numerical Answer Type:
Enter a numerical valueAnswer: 1 to 1+4 marks
Solution & Explanation
Related Formula
Magnetic field at the center of a circular segment of angle θ$\theta$:
Let's analyze the contributions from each part of the loop to the magnetic field at the center O$O$:
The straight wire segments AB$AB$ and CD$CD$ lie along radial lines passing directly through O$O$. Since d l ∥ r$d\vec{l} \parallel \vec{r}$, their magnetic field contribution is zero:
BAB = BCD = 0$$B_{\text{AB}} = B_{\text{CD}} = 0$$
Semicircular segment of radius R₂ = 4π~m$R_2 = 4\pi\mathrm{~m}$ carries current creating a field pointing out of the page (by right-hand rule):
BR2 = (μ₀ I)/(4 R₂) (Out of page)$$B_{R2} = \frac{\mu_0 I}{4 R_2} \quad (\text{Out of page})$$
Semicircular segment of radius R₁ = 6π~m$R_1 = 6\pi\mathrm{~m}$ carries current creating a field pointing into the page:
Comparing this to k × 10⁻⁷~T$k \times 10^{-7}\mathrm{~T}$:
k = 1$k = 1$
Pattern Recognition
Notice how the radial straight wires never contribute to the magnetic field at the center. Semicircles with opposite currents simply subtract. Symmetrizing the math early by factoring out (μ₀ I)/(4π)$\frac{\mu_0 I}{4\pi}$ makes the calculations incredibly neat and quick.
Chapter Mix
Class 12 Physics: Magnetic Effects of Current: Biot-Savart Law
Keywords:#Biot Savart concentric semicircles#magnetic field circular segment#JEE Main 2025 Morning Q24#resultant magnetic field concentric#magnetic field loop#semicircular segments#Biot Savart law
More Magnetic Effects of Current Previous-Year Questions — Page 6
Qjee_main_2025_28_jan_eveningBiot Savart Law
An infinite wire has a circular bend of radius a$a$ , and carrying a current I as shown in figure. The magnitude of magnetic field at the origin O of the arc is given by:
An infinite current carrying wire presenting a three-quarter circular bend around center origin O.
Let us decompose the structure into three functional parts as mapped out below:
An infinite current carrying wire presenting a three-quarter circular bend around center origin O.
Segment 1 (Incoming semi-infinite line): The straight line extends to infinity, with its terminating tip at a perpendicular distance a$a$ from origin O$O$. Using the right-hand grip rule, the direction points into the page:
Segment 2 (Three-quarter circular loop): The loop forms an angle of θ = (3π)/(2)$\theta = \frac{3\pi}{2}$ radians around O$O$. The field points into the page:
Always check the axis alignment first. Any straight wire segment whose extended line passes directly through the field point contributes exactly zero to the total magnetic field value.
Qjee_main_2025_29_jan_morningAmpere\'s Circuital Law
Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire\'s cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be
A galvanometer has a resistance of 50~Ω$50\mathrm{~\Omega}$ and it allows maximum current of 5~mA$5\mathrm{~mA}$. It can be converted into voltmeter to measure upto 100~V$100\mathrm{~V}$ by connecting in series a resistor of resistance:
Voltmeter resistance is always high because it is connected in parallel to circuits to prevent current drawing leaks.
Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Class 12 Physics: Current Electricity
Qjee_main_2024_01_february_morningMagnetic Field due to a Current Element
A regular polygon of 6 sides is formed by bending a wire of length 4pi$4pi$ meter. If an electric current of 4pisqrt3~A$4pisqrt{3}\mathrm{~A}$ is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be x × 10⁻⁷~T$x \times 10^{-7}\mathrm{~T}$. The value of x$x$ is:
Numerical Answer.Answer: 72 to 72
Solution
Related Formula
Magnetic field due to a straight wire segment of length 2L$2L$ at distance r$r$:
Total perimeter = 6 · a = 4π side length a = (4π)/(6) = (2π)/(3)~m$= 6 \cdot a = 4\pi \implies \text{side length } a = \frac{4\pi}{6} = \frac{2\pi}{3}\mathrm{~m}$.
For a regular hexagon segment, the interior angles relative to the normal vector are θ₁ = θ₂ = 30^°$\theta_1 = \theta_2 = 30^\circ$.
The normal distance r$r$ from the center to a side is:
For regular polygons, the normal distance r$r$ to the side is always r = (a)/(2) ((π)/(n))$r = \frac{a}{2}\cot(\frac{\pi}{n})$. The contribution from all n$n$ symmetric segments adds up constructively.
Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q39jee_main_2024_29_january_eveningMotion of Charged Particle in Magnetic Field
Two particles X and Y having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii R₁$R_1$ and R₂$R_2$ respectively. The mass ratio of X and Y is:
A.((R₂)/(R₁))²$\left(\frac{R_2}{R_1}\right)^2$
B.((R₁)/(R₂))²$\left(\frac{R_1}{R_2}\right)^2$
C.((R₁)/(R₂))$\left(\frac{R_1}{R_2}\right)$
D.((R₂)/(R₁))$\left(\frac{R_2}{R_1}\right)$
Solution
Related Formula
The radius R$R$ of the path of a charged particle moving perpendicular to a magnetic field B$B$ is:
R = (mv)/(qB) = (p)/(qB)$$R = \frac{mv}{qB} = \frac{p}{qB}$$
In terms of kinetic energy K$K$:
R = √(2mK)qB$$R = \frac{\sqrt{2mK}}{qB}$$
Since the particle is accelerated through potential V$V$, kinetic energy K = qV$K = qV$:
Shortcut: Whenever charges and potential differences are equal, the radius of orbit in a magnetic field scales as R ∝ √(m)$R \propto \sqrt{m}$. Squaring both sides yields m ∝ R²$m \propto R^2$ instantly.
Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
More Magnetic Effects of Current Questions — jee_main_2025_03_april_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
