NEET UG 2019 — Physics Electromagnetism

The magnetic field at the centre of an arc subtended at an angle $\theta$θ is given by $\mathrm{B}=\frac{\mu_0 \mathrm{i}}{2 \mathrm{R}} \times \frac{\theta}{2 \pi}$B=2Rμ0​i​×2πθ​ Then, the magnetic field due to larger arc $A B$AB is $\mathrm{B}_1=\frac{\mu_0 \mathrm{i}_1}{2 \mathrm{R}} \times \frac{270}{2 \pi}$B1​=2Rμ0​i1​​×2π270​ which acts in inward direction according to right hand thumb rule. And magnetic field due to smaller arc $A B$AB is $\mathrm{B}_2=\frac{\mu_0 \mathrm{i}_2}{2 \mathrm{R}} \times \frac{90}{2 \pi}$B2​=2Rμ0​i2​​×2π90​ which acts in outward direction. The resultant magnetic field $$\begin{aligned} & \mathrm{B}_{\mathrm{R}}=\mathrm{B}_1+\mathrm{B}_2 \\ & =-\frac{\mu_0 \mathrm{i}_1 \times 270}{4 \pi \mathrm{R}}+\frac{\mu_0 \mathrm{i}_2 \times 90}{4 \pi \mathrm{R}} \text { [From Eq. (i) and (ii)] } \end{aligned}$$​BR​=B1​+B2​=−4πRμ0​i1​×270​+4πRμ0​i2​×90​ [From Eq. (i) and (ii)] ​ which acts in inward direction as $\mathrm{B}_1>\mathrm{B}_2$B1​>B2​. Two arcs can also be seen as the two resistances in parallel combination. So, the potential across them will be same i.e. $$\begin{aligned} \mathrm{V}_1 & =\mathrm{V}_2 \\ \mathrm{i}_1 \mathrm{R}_1 & =\mathrm{i}_2 \mathrm{R}_2 \end{aligned}$$V1​i1​R1​​=V2​=i2​R2​​ where, $\mathrm{R}_1$R1​ and $\mathrm{R}_2=$R2​= Resistance of respective segments The wire is uniform so $$\begin{aligned} & \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{\mathrm{R} \times 270}{\mathrm{R} \times 90} [\because \text { length of arc }=\text { radius } \times \\ & \text { angle] } \end{aligned}$$​R2​R1​​= L2​L1​​=R×90R×270​[∵ length of arc = radius × angle] ​ From Eq. (iv), we get $\Rightarrow \frac{\mathrm{i}_1}{\mathrm{i}_2}=\frac{\mathrm{R}_2}{\mathrm{R}_1}=\frac{90}{270}=\frac{1}{3}$⇒i2​i1​​=R1​R2​​=27090​=31​ or $3 i_1=i_2$3i1​=i2​ From Eq. (iii) and (v), we get $$\begin{aligned} \mathrm{B}_{\mathrm{R}} & =\frac{\mu_0}{4 \pi \mathrm{R}}\left(-270 \mathrm{i}_1+90 \mathrm{i}_2\right) \\ & =\frac{\mu_0}{4 \pi R}\left[-270 \mathrm{i}_1+90\left(3 \mathrm{i}_1\right)\right] \\ & =\frac{\mu_0}{4 \pi \mathrm{R}}\left(-270 \mathrm{i}_1+270 \mathrm{i}_1\right)=0 \end{aligned}$$BR​​=4πRμ0​​(−270i1​+90i2​)=4πRμ0​​[−270i1​+90(3i1​)]=4πRμ0​​(−270i1​+270i1​)=0​