Question

2 long parallel wires in yz plane at a distance 2a apart carry a steady current I in opposite drections. midway there isa. rectangular loops of weire 2b x d carrying current I as shown, b<a<2b, the loops is free to rotate about z axisand currents remin fixed to I irrespective of relative motion between loop and wire. phi is acute angle that plane of loop makes with plane of wire. torque tending to rotate looo abut its axis is max for sin phi=?

Answer 1

$\sin\phi=\frac{a}{\sqrt{2a^2-b^2}}$

Answer 2

We will show that after some algebra the torque on the rectangular loop is maximum when

$$\sin\phi=\frac{a}{\sqrt{2a^2-b^2}}.$$

Below is one way to arrive at the answer.

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Step‐by‐step solution (minimal explanation):

1. Geometry & Setup:
    – Two long wires lie in the $yz$–plane at $x=\pm a$ carrying current $I$ in opposite directions.
    – A rectangular loop of dimensions $2b \times d$ (with $2b$ along the horizontal direction) is centered midway between them.
    – The loop is free to rotate about the $z$–axis; its plane makes an acute angle $\phi$ with the $yz$–plane (the plane of the wires).
    – Due to the rotation, the vertical sides of the loop (carrying current $I$ over a length $d$) are displaced horizontally by $b\sin\phi$ from the $z$–axis, thereby “feeling” different magnetic fields from the two wires.

2. Force & Torque:
    – The magnetic field due to a long wire at a perpendicular distance $r$ is

$$B=\frac{\mu_0 I}{2\pi r}.$$

 – Each vertical side experiences a force $F=Id\,B$ with the difference of the magnitudes coming from the different distances $a\mp b\sin\phi$ from the two wires.
 – Since the two forces act at distances $b\cos\phi$ (the lever arm) from the rotation axis, one may show that the net torque (apart from overall constant factors) may be written as

$$\tau\propto\frac{\sin\phi\cos\phi}{a^2-b^2\sin^2\phi}.$$

3. Maximizing the Torque:
    – Differentiate the expression

$$f(\phi)=\frac{\sin\phi\cos\phi}{a^2-b^2\sin^2\phi}$$

 with respect to $\phi$ and set the derivative to zero. After some algebra (and using the double‐angle identities $\cos2\phi=2\cos^2\phi-1$ and $\sin^2\phi=\frac{1-\cos2\phi}{2}$) one obtains the condition

$$a^2\cos2\phi+b^2\sin^2\phi=0.$$

 – Rewriting $\sin^2\phi=\frac{1-\cos2\phi}{2}$ we get

$$a^2\cos2\phi+\frac{b^2}{2}(1-\cos2\phi)=0$$

which leads to

$$\cos2\phi=-\frac{b^2}{2a^2-b^2}.$$

4. Expressing in Terms of $\sin\phi$:
    – Using the identity $\cos2\phi=1-2\sin^2\phi$ we have

$$1-2\sin^2\phi=-\frac{b^2}{2a^2-b^2}.$$

 – Solving for $\sin^2\phi$ we find

$$2\sin^2\phi=1+\frac{b^2}{2a^2-b^2}\quad\Longrightarrow\quad \sin^2\phi=\frac{a^2}{2a^2-b^2}.$$

Thus,

$$\sin\phi=\frac{a}{\sqrt{2a^2-b^2}},$$

where we take the positive square root since $\phi$ is acute.

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Summary of Answer:

$$\boxed{\sin\phi=\frac{a}{\sqrt{2a^2-b^2}}.}$$

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Additional details:

• Subject: Physics
• Chapter: Current Electricity (in NCERT Class 12 Physics)
• Topic: Force on a current-carrying conductor in a magnetic field; Torque on a current loop.
• Difficulty level: Medium
• Question type: single_choice (if options were given) or descriptive

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This solution uses ideas familiar to JEE/NEET students: evaluating the net force on a differential element of a loop in a magnetic field produced by long wires and then maximizing the resulting torque. The key step is to write the torque (proportional to $\sin\phi\cos\phi/(a^2-b^2\sin^2\phi)$) and set its derivative with respect to $\phi$ equal to zero.