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Question #43335

Difficulty: 8/10 | 🟢 Free

The Question

A rigid rectangular coil of NN turns, area AA, and mass mm is hinged along its top horizontal edge (aligned with the zz-axis) and hangs vertically in the yzyz-plane under uniform gravity gg (acting in the −y-y direction). The coil carries an alternating current I(t)=I0sin⁡(ωt)I(t) = I_0 \sin(\omega t). A uniform horizontal magnetic field B⃗=B0i^\vec{B} = B_0 \hat{i} is present in the region. The hinge is attached to a cart that accelerates horizontally in the +x+x direction with an acceleration a⃗(t)=a0sin⁡(ωt)i^\vec{a}(t) = a_0 \sin(\omega t) \hat{i}. Assume the angular frequency ω\omega is much smaller than the natural frequency of the coil as a pendulum, so the coil continuously maintains its quasi-static equilibrium position relative to the non-inertial frame of the cart. Also assume the magnetic torque is negligible in determining the angular position of the coil, and a0≪ga_0 \ll g so small angle approximations are valid. What is the magnitude of the time-averaged magnetic torque on the coil over one complete cycle?

Options

  • A.

    NI0AB0a02g\frac{N I_0 A B_0 a_0}{2 g}

  • B.

    00

  • C.

    NI0AB0a0g\frac{N I_0 A B_0 a_0}{g}

  • D.

    NI0AB0g2a0\frac{N I_0 A B_0 g}{2 a_0}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Wattless Current|Torque on Current Loop|Non-Inertial Frame|Vector Cross-Product

Why This Question Is Tricky

Common Mistakes

This question tests the misconception that a 'wattless' AC current (which averages to zero) cannot produce a net time-averaged mechanical interaction.

Students typically assume ⟨τ⟩=⟨μ(t)×B⟩=0\langle \tau \rangle = \langle \mu(t) \times B \rangle = 0 because the current is purely sinusoidal.

However, they fail to account for the non-inertial frame: the pseudo-force causes the coil's orientation θ(t)\theta(t) to oscillate in phase with the current.

This creates a non-linear coupling where the instantaneous Lorentz force depends on sin⁡2(ωt)\sin^2(\omega t), leading to a non-zero time-averaged torque.

The trap is choosing zero (Option B) by ignoring the cross-product of the time-varying pseudo-force-induced moments and the magnetic moment.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterAlternating Current - Class 12
ConceptWattless Current
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree