Correct Answer is Option (B)

For a current carrying wire, magnetic field at a distance r is given by


$B=\frac{{\mu }_{0}i}{4\pi r}(\sin {\theta }_1+\sin {\theta }_2)$

Now, in given case,



Due to symmetry of arrangement, net field at centre of triangle is

$\begin{array}{rl}{B}_{\text{net }}& =\text{ Sum of fields of all wires (sides) }\\ \\ & =3\times \frac{{\mu }_{0}i}{4\pi r}(\sin {\theta }_1+\sin {\theta }_2)\end{array}$

Here, ${\theta }_1={\theta }_2={60}^{\circ }$

$\begin{array}{rl}& \therefore \sin {\theta }_1=\sin {\theta }_2=\frac{\sqrt{3}}{2},i=10\mathrm{A},\frac{{\mu }_0}{4\pi }={10}^{-7}{\mathrm{NA}}^{-2}\\ \\ & \text{ and }r=\frac{1}{3}\times \text{ altitude }\\ \\ & =\frac{1}{3}\times \frac{\sqrt{3}}{2}\times \text{ sides length }=\frac{1}{2\sqrt{3}}\times 1\mathrm{m}=\frac{1}{2\sqrt{3}}\mathrm{m}\end{array}$

So,

$\begin{array}{rl}{B}_{\mathrm{net}}& =\frac{3\times {10}^{-7}\times 10\times 2\left(\frac{\sqrt{3}}{2}\right)}{\left(\frac{1}{2\sqrt{3}}\right)}=18\times {10}^{-6}\mathrm{T}\\ \\ \Rightarrow B_{\mathrm{net}}& =18\mu \mathrm{T}\end{array}$

Correct Answer is Option (C)

Magnetic field at point P

${\overrightarrow{B}}_{net}=\frac{{\mu }_{0}i}{2\pi d}(-\hat{k})+\frac{{\mu }_{0}i}{2\pi d}(\hat{k})=0$

Correct Answer is Option (D)

For loop,

L = 2$\pi$R

For coil,

L = N $\times$ 2$\pi$r

$\therefore$   2$\pi$R = N $\times$ 2$\pi$r

$\Rightarrow$  R = Nr

$\Rightarrow$  r = $\frac{R}{N}$

We know,

B_L = $\frac{{\mu }_{0}i}{2R}$

and  B_C = N $\times$ $\frac{{\mu }_{0}i}{2r}$

$\therefore$  $\frac{B_L}{B_C}=\frac{\frac{{\mu }_{0}i}{2R}}{N\times \frac{{\mu }_{0}i}{2\left(\frac{R}{N}\right)}}=\frac{1}{N^2}$

Correct Answer is Option (D)

Magnetic field due to circular arc

$\overrightarrow{B}=\frac{{\mu }_{0}i\theta }{4\pi r}$

Due to QR arc magnetic field is outward direction of the plane.

And due to PS arc magnetic field is inward direction of the plane,

So, net magnetic field,

$\overrightarrow{B}=\left({\overrightarrow{B}}_{QR}-{\overrightarrow{B}}_{PS}\right)\hat{K}$

$\overrightarrow{B}=\frac{{\mu }_{0}i\theta }{4\pi }\left(\frac{1}{3\times {10}^{-2}}-\frac{1}{5\times {10}^{-2}}\right)\hat{K}$

$\overrightarrow{B}={10}^{-7}\times 10\times \frac{\pi }{4}\left(\frac{1}{3\times {10}^{-2}}-\frac{1}{5\times {10}^{-2}}\right)\hat{K}$

$\left|\overrightarrow{B}\right|=1\times {10}^{-5}T$

Correct Answer is Option (D)

We know that magnetic field due to finite current carrying wire is

$B=\frac{{\mu }_0}{4\pi }\frac{I}{b}(\cos {\theta }_1+\cos {\theta }_2)$ ..... (1)

Given that side of triangle = 4.5 $\times$ 10^$-$2 m = a; current = 1A since the triangle is equilateral, angle of each side will be 60${}^{\circ }$.

Now, $\tan \theta =\frac{Perpendicular}{Base}\Rightarrow \tan {60}^{\circ }=\frac{a/2}{b}$

$\Rightarrow b=\frac{a}{2\tan {60}^{\circ }}=\frac{a}{2\sqrt{3}}$

Using equation (1), we get

$B=\frac{{\mu }_0}{4\pi }\frac{I}{a/2\sqrt{3}}(\cos {30}^{\circ }+\cos {30}^{\circ })=\frac{{\mu }_0}{4\pi }\frac{2\sqrt{3}I}{a}2\cos {30}^{\circ }$

$B=\frac{{\mu }_0}{4\pi }\frac{2\sqrt{3}I}{a}\frac{2\times \sqrt{3}}{2}=\frac{{\mu }_0}{4\pi }\frac{6I}{a}$

$\Rightarrow B=\frac{{10}^{-7}\times 6\times 1}{4.5\times {10}^{-2}}=1.33\times {10}^{-5}\sim 2\times {10}^{-5}$ Wb/m²