A wire in the form of a square of side a carr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The magnetic field produced by a single side of the square (e.g.,side $AB$) at the centre is given by the formula for a finite wire: $B_1 = \frac{

Vedclass · Accepted Answer

$\frac{2 \sqrt{2} \mu_0 i}{\pi a}$

Vedclass · Answer

(C) The magnetic field produced by a single side of the square (e.g.,side $AB$) at the centre is given by the formula for a finite wire: $B_1 = \frac{\mu_0 i}{4 \pi r} (\sin \phi_1 + \sin \phi_2)$.Here, the distance from the centre to the side is $r = \frac{a}{2}$, and the angles subtended at the centre by the ends of the side are $\phi_1 = 45^{\circ}$ and $\phi_2 = 45^{\circ}$.Substituting these values: $B_1 = \frac{\mu_0 i}{4 \pi (a/2)} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{\mu_0 i}{2 \pi a} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 i}{2 \pi a} (\frac{2}{\sqrt{2}}) = \frac{\sqrt{2} \mu_0 i}{2 \pi a} = \frac{\mu_0 i}{\sqrt{2} \pi a}$.Since there are $4$ identical sides, the total magnetic field at the centre is $B = 4 B_1 = 4 	imes \frac{\mu_0 i}{2 \sqrt{2} \pi a} = \frac{2 \sqrt{2} \mu_0 i}{\pi a}$.

$(i)$	$(ii)$	$(iii)$
$(A). \frac{\mu_0 i}{r} \otimes$	$(A). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \otimes$	$(A). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \otimes$
$(B). \frac{\mu_0 i}{2r} \odot$	$(B). \frac{\mu_0 i}{4}(\frac{1}{r_1} + \frac{1}{r_2}) \otimes$	$(B). \frac{\mu_0 i}{4}(\frac{1}{r_1} + \frac{1}{r_2}) \otimes$
$(C). \frac{\mu_0 i}{4r} \otimes$	$(C). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \odot$	$(C). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \odot$
$(D). \frac{\mu_0 i}{4r} \odot$	$(D). 0$	$(D). 0$

$A$ wire in the form of a square of side $a$ carries a current $i$. Then, the magnetic induction at the centre of the square is (Magnetic permeability of free space $= \mu_0$)

Similar Questions

Two insulated circular loops $A$ and $B$ of radius '$a$' carry a current of '$I$' in the directions as shown in the figure. The magnitude of the magnetic induction at the centre $O$ will be:

Do magnetic forces obey Newton's third law? Verify for two current elements $\overrightarrow{dl_1} = dl(\hat{i})$ located at the origin and $\overrightarrow{dl_2} = dl(\hat{j})$ located at $(0, R, 0)$. Both carry current $I$.

An infinitely long straight conductor is bent into the shape as shown in the figure. It carries a current $I$ and the radius of the circular loop is $r$. The magnetic induction at the center $O$ of the circular loop is:

Two long straight parallel wires $A$ and $B$ separated by $5 \ m$ carry currents $2 \ A$ and $6 \ A$ respectively in the same direction. The resultant magnetic field due to the two wires at a point $P$ at a distance of $2 \ m$ from wire $A$ in between the two wires is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module