Electromagnetic Induction: Step-by-Step Worked Examples (2)

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Tags Electromagnetic Induction

Question

A rectangular loop of length L=0.2L = 0.2 m and width w=0.1w = 0.1 m moves with velocity v=5v = 5 m/s in a uniform magnetic field B=0.4B = 0.4 T perpendicular to the loop’s plane. The loop has resistance R=2 ΩR = 2\,\Omega. Find the induced EMF, the current, and the force needed to maintain the motion as the loop enters the field region.

Solution — Step by Step

Only the edge of the loop crossing the field boundary cuts flux lines as the loop enters. That edge has length L=0.2L = 0.2 m. The motional EMF is ε=BLv\varepsilon = B L v.

ε=BLv=0.4×0.2×5=0.4 V\varepsilon = B L v = 0.4 \times 0.2 \times 5 = 0.4 \text{ V}

I=εR=0.42=0.2 AI = \frac{\varepsilon}{R} = \frac{0.4}{2} = 0.2 \text{ A}

The current-carrying edge inside the field experiences a force F=BILF = B I L opposing motion (Lenz’s law). To maintain constant velocity, the external agent must apply an equal and opposite force:

Fext=BIL=0.4×0.2×0.2=0.016 NF_{\text{ext}} = B I L = 0.4 \times 0.2 \times 0.2 = 0.016 \text{ N}

Why This Works

Motional EMF ε=BLv\varepsilon = BLv comes from the magnetic force on free charges in the moving conductor. Once current flows, that current also experiences a force due to the field — by Lenz’s law, this force always opposes the cause (the motion). So an external agent has to do positive work to keep the loop moving.

The energy balance: Pext=Fextv=0.016×5=0.08P_{\text{ext}} = F_{\text{ext}} v = 0.016 \times 5 = 0.08 W. Power dissipated in resistor: PR=I2R=0.04×2=0.08P_R = I^2 R = 0.04 \times 2 = 0.08 W. The two match — energy is conserved.

Whenever the question asks “force to maintain motion,” compute I2R/vI^2 R / v as a quick check. If your FextF_{\text{ext}} doesn’t match PR/vP_R / v, you’ve made an error.

Alternative Method

Use flux directly. As the loop enters the field, flux through the loop increases at rate dΦdt=BLv\dfrac{d\Phi}{dt} = B L v. By Faraday’s law ∣ε∣=dΦ/dt=BLv|\varepsilon| = d\Phi/dt = BLv. Same answer.

Students multiply by the loop’s width ww instead of length LL. The formula uses the length of the edge perpendicular to velocity that is cutting field lines. Always identify that edge first.

Final answer: ε=0.4\varepsilon = 0.4 V, I=0.2I = 0.2 A, Fext=0.016F_{\text{ext}} = 0.016 N.

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