There are two infinitely long straight curren | vedclass

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

Question

${B}_{	ext {due to wire }(1)}=\frac{\mu_{{o}} {I}}{4 \pi {y}}\left[\sin 90+\sin 	heta_{1}ight]$

$=\frac{\mu_{0}}{4 \pi} \frac{{I}}{{y}}\left(1+

Vedclass · Accepted Answer

$\frac{\mu_{0} I}{4 \pi x y}\left[\sqrt{x^{2}+y^{2}}+(x+y)\right]$

Vedclass · Answer

${B}_{	ext {due to wire }(1)}=\frac{\mu_{{o}} {I}}{4 \pi {y}}\left[\sin 90+\sin 	heta_{1}ight]$

$=\frac{\mu_{0}}{4 \pi} \frac{{I}}{{y}}\left(1+\frac{{x}}{\sqrt{{x}^{2}+{y}^{2}}}ight) \ldots \ldots(1)$

${B}_{	ext {due to wire }(2)}=\frac{\mu_{0}}{4 \pi} \frac{{I}}{{x}}\left(\sin 90^{\circ}+\sin 	heta_{2}ight)$

$=\frac{\mu_{0}}{4 \pi} \frac{{I}}{{x}}\left(1+\frac{{y}}{\sqrt{{x}^{2}+{y}^{2}}}ight) \ldots \ldots(2)$

Total magnetic field

${B}={B}_{1}+{B}_{2}$

${B}=\frac{\mu_{0} {I}}{4 \pi}\left[\frac{1}{{y}}+\frac{{x}}{{y} \sqrt{{x}^{2}+{y}^{2}}}+\frac{1}{{x}}+\frac{{y}}{{x} \sqrt{{x}^{2}+{y}^{2}}}ight]$

${B}=\frac{\mu_{0} {I}}{4 \pi}\left[\frac{{x}+{y}}{{xy}}+\frac{{x}^{2}+{y}^{2}}{{xy} \sqrt{{x}^{2}+{y}^{2}}}ight]$

${B}=\frac{\mu_{0} {I}}{4 \pi}\left[\frac{{x}+{y}}{{xy}}+\frac{\sqrt{{x}^{2}+{y}^{2}}}{{xy}}ight]$

${B}=\frac{\mu_{0} {I}}{4 \pi {xy}}\left[\sqrt{{x}^{2}+{y}^{2}}+({x}+{y})ight]$

There are two infinitely long straight current carrying conductors and they are held at right angles to each other so that their common ends meet at the origin as shown in the figure given below. The ratio of current in both conductor is $1: 1$. The magnetic field at point $P$ is ...... .

Similar Questions

Tesla is the unit of

A straight wire carrying a current $10\, A$ is bent into a semicircular arc of radius $5\, cm.$ The magnitude of magnetic field at the center is

Gauss is unit of which quantity

Figure shows the cross-sectional view of the hollow cylindrical conductor with inner radius '$R$' and outer radius '$2R$', cylinder carrying uniformly distributed current along it's axis. The magnetic induction at point '$P$' at a distance $\frac{{3R}}{2}$ from the axis of the cylinder will be

Charge $q$ is uniformly spread on a thin ring of radius $R.$ The ring rotates about its axis with a uniform frequency $f\, Hz.$ The magnitude of magnetic induction at the center of the ring is