Question:
A square loop of side \(a\) carries current \(i\). Find magnetic field at centre.
A square loop of side \(a\) carries current \(i\). Find magnetic field at centre.
Show Hint
Square loop $\Rightarrow$ calculate one side field and multiply by 4.
Updated On: Apr 23, 2026
- \(\frac{\mu_0 i}{2\pi a}\)
- \(\frac{\mu_0 i\sqrt{2}}{\pi a}\)
- \(\frac{2\sqrt{2}\mu_0 i}{\pi a}\)
- \(\frac{\mu_0 i}{\sqrt{2}\pi a}\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
Magnetic field due to finite wire:
\[
B = \frac{\mu_0 i}{4\pi r}(\sin\theta_1 + \sin\theta_2)
\]
Step 1: Geometry
Distance from centre to side: \[ r = \frac{a}{2} \] \[ \theta = 45^\circ \]
Step 2: Field due to one side
\[ B_1 = \frac{\mu_0 i}{4\pi (a/2)}(2\sin45^\circ) \] \[ = \frac{\mu_0 i}{2\pi a}(2 \cdot \frac{1}{\sqrt{2}}) = \frac{\mu_0 i\sqrt{2}}{\pi a} \]
Step 3: Total field (4 sides)
\[ B = 4B_1 = \frac{2\sqrt{2}\mu_0 i}{\pi a} \] Conclusion: \[ \frac{2\sqrt{2}\mu_0 i}{\pi a} \]
Step 1: Geometry
Distance from centre to side: \[ r = \frac{a}{2} \] \[ \theta = 45^\circ \]
Step 2: Field due to one side
\[ B_1 = \frac{\mu_0 i}{4\pi (a/2)}(2\sin45^\circ) \] \[ = \frac{\mu_0 i}{2\pi a}(2 \cdot \frac{1}{\sqrt{2}}) = \frac{\mu_0 i\sqrt{2}}{\pi a} \]
Step 3: Total field (4 sides)
\[ B = 4B_1 = \frac{2\sqrt{2}\mu_0 i}{\pi a} \] Conclusion: \[ \frac{2\sqrt{2}\mu_0 i}{\pi a} \]
Was this answer helpful?
0
0
Top MET Physics Questions
- A coil of 100 turns and area $2 \times 10^{-2}m^2 $ is pivoted about a vertical diameter in a uniform magnetic field and carries a current of 5A. When the coil is held with its plane in north-south direction, it experiences a couple of 0.33 Nm. When the plane is east-west, the corresponding couple is 0.4 Nm, the value of magnetic induction is [Neglect earth's magnetic field]
- Radiation, with wavelength 6561 $?$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4} T$. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :
- In intrinsic semiconductor at room temperature number of electrons and holes are
- A rope of mass $0.1\, kg$ is connected at the same height of two opposite walls. It is allowed to hang under its own weight. At [he contact point between the rope and the wall, the rope makes an angle $ \theta =10^{\circ} $ with respect to horizontal. The tension in the rope at its midpoint between the walls is
- At great distances from an electric dipole, the electric field strength due to the dipole varies with the distance as
View More Questions
Top MET Magnetic Field Questions
Top MET Questions
- A coil of 100 turns and area $2 \times 10^{-2}m^2 $ is pivoted about a vertical diameter in a uniform magnetic field and carries a current of 5A. When the coil is held with its plane in north-south direction, it experiences a couple of 0.33 Nm. When the plane is east-west, the corresponding couple is 0.4 Nm, the value of magnetic induction is [Neglect earth's magnetic field]
- $\int\limits_{0}^{8}| x -5| d x$ is equal to
- Radiation, with wavelength 6561 $?$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4} T$. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :
- In intrinsic semiconductor at room temperature number of electrons and holes are
- Which of the following product is formed in the reaction $ CH_3MgBr {->[(i)CO_2][(ii)H_2O]} ? $
View More Questions