JEE Advanced 2023 Paper 2 · Q09 · Biot-Savart Law
A rectangular conducting loop of length $4$ cm and width $2$ cm is in the $xy$-plane, as shown in the figure. It is being moved away from a thin and long conducting wire along the direction $\dfrac{\sqrt{3}}{2}\hat{x} + \dfrac{1}{2}\hat{y}$ with a constant speed $v$. The wire is carrying a steady current $I = 10$ A in the positive $x$-direction. A current of $10\,\mu A$ flows through the loop when it is at a distance $d = 4$ cm from the wire. If the resistance of the loop is $0.1\,\Omega$, then the value of $v$ is ___ m·s$^{-1}$. [Given: $\mu_0 = 4\pi\times 10^{-7}$ N·A$^{-2}$]
Reveal answer + step-by-step solution
Correct answer:4
Solution
Loop sides parallel to the wire have length $\ell=2$ cm; the perpendicular sides have length $4$ cm. The near edge is at $d=4$ cm from the wire, the far edge at $d+4$ cm $= 8$ cm.
Field from the long wire at distance $r$: $B(r)=\mu_{0}I/(2\pi r)$. $$B_{\text{near}} = \dfrac{(4\pi\times 10^{-7})(10)}{2\pi (0.04)} = 5\times 10^{-5}\ \text{T},\quad B_{\text{far}} = \dfrac{\mu_{0}I}{2\pi(0.08)} = 2.5\times 10^{-5}\ \text{T}.$$
Only the velocity component perpendicular to the wire (i.e. along $\hat y$) drives flux change. $\vec v = v\!\left(\tfrac{\sqrt 3}{2}\hat x + \tfrac{1}{2}\hat y\right)$, so $v_{y}=v/2$.
Net motional EMF: $\varepsilon = (B_{\text{near}}-B_{\text{far}})\,\ell\,v_{y} = (2.5\times 10^{-5})(0.02)(v/2) = 2.5\times 10^{-7}\,v$ V.
Current: $i = \varepsilon/R = 2.5\times 10^{-6}\,v$. Setting $i = 10\times 10^{-6}$ A gives $\boxed{v=4}$ m/s.
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