Figure shows a current carrying square loop ABCD of edge length is $ a $ lying in a plane. If the resistance of the ABC part is $ r $ and that of the ADC part is $ 2r $, then the magnitude of the resultant magnetic field at the center of the square loop is: 
Figure shows a current carrying square loop ABCD of edge length is $ a $ lying in a plane. If the resistance of the ABC part is $ r $ and that of the ADC part is $ 2r $, then the magnitude of the resultant magnetic field at the center of the square loop is:
Show Hint
\( \frac{\sqrt{2}\mu_0 I}{3 \pi a} \)
- \( \frac{\mu_0 I}{2 \pi a} \)
- \( \frac{2 \mu_0 I}{3 \pi a} \)
- \( \frac{3 \pi \mu_0 I}{\sqrt{2}} \)
The Correct Option is A
Approach Solution - 1
Given the diagram, we have the following information:
\( R_{ABC} = r \), and \( R_{ADC} = 2r \)
Also, the currents \( i_1 \) and \( i_2 \) are given by:
\( i_1 = \frac{2I}{3} \), and \( i_2 = \frac{I}{3} \)
The magnetic field at the center \( B_{\text{centre}} \) is calculated as:
\( B_{\text{centre}} = \frac{2\mu_0 I \sqrt{2}}{4 \pi \left( \frac{a}{2} \right)} \left[ \frac{2I}{3} - \frac{I}{3} \right] = \sqrt{2} \frac{\mu_0 I}{3\pi a} \)
Approach Solution -2
The magnetic field at the center of a square loop is given by the formula: \[ B = \frac{\mu_0 I}{2a} \left( \frac{2}{\pi} \right) \] However, the presence of different resistances for parts ABC and ADC requires us to calculate the net effective current flowing through each section and the resultant magnetic field produced.
For each segment, we calculate their individual contributions based on their respective resistances, and by applying the Biot-Savart law, we arrive at the magnetic field at the center of the loop as: \[ B = \frac{\sqrt{2} \mu_0 I}{3 \pi a} \] Thus, the correct answer is Option 1.
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Electromagnetism Questions
- Match List I with List II
List I List II A. \( \oint \vec{B} \cdot d\vec{l} = \mu_0 i_c + \mu_0 \epsilon_0 \frac{d\phi_E}{dt} \) I. Gauss' law for electricity B. \( \oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt} \) II. Gauss' law for magnetism C. \( \oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0} \) III. Faraday law D. \( \oint \vec{B} \cdot d\vec{A} = 0 \) IV. Ampere – Maxwell law
Choose the correct answer from the options given below - Given below are two statements:
Statement (I): When currents vary with time, Newton’s third law is valid only if momentum carried by the electromagnetic field is taken into account.
Statement (II): Ampere’s circuital law does not depend on Biot-Savart’s law.
In the light of the above statements, choose the correct answer from the options given below: - An electron with kinetic energy \(5 \, \text{eV}\) enters a region of uniform magnetic field of \(3 \, \mu\text{T}\) perpendicular to its direction. An electric field \(E\) is applied perpendicular to the direction of velocity and magnetic field. The value of \(E\), so that the electron moves along the same path, is ______ \( \text{NC}^{-1} \).
Given: mass of electron = \(9 \times 10^{-31} \, \text{kg}\), electric charge = \(1.6 \times 10^{-19} \, \text{C}\) - A square loop PQRS having 10 turns, area \(3.6 \times 10^{-3} \, \text{m}^2\), and resistance \(100 \, \Omega\) is slowly and uniformly being pulled out of a uniform magnetic field of magnitude \(B = 0.5 \, \text{T}\) as shown. Work done in pulling the loop out of the field in \(1.0 \, \text{s}\) is ______ \( \times 10^{-6} \, \text{J} \).
- A galvanometer of resistance \( 100 \, \Omega \) when connected in series with \( 400 \, \Omega \) measures a voltage of up to \( 10 \, V \). The value of resistance required to convert the galvanometer into an ammeter to read up to \( 10 \, A \) is \( x \times 10^{-2} \, \Omega \). The value of \( x \) is:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.