Electromagnetic Induction: Application Problems (3)

hard • 3 min read
Tags Electromagnetic Induction

Question

A conducting rod of length L=0.5 mL = 0.5 \text{ m} slides with constant velocity v=4 m/sv = 4 \text{ m/s} on two parallel frictionless rails connected by a 2Ω2 \Omega resistor. The setup is in a uniform magnetic field B=0.5 TB = 0.5 \text{ T} perpendicular to the plane of the rails. The rod itself has resistance 0.5Ω0.5 \Omega. Find (a) the EMF induced, (b) the current through the resistor, (c) the force needed to keep the rod moving at constant velocity, and (d) the power dissipated.

Solution — Step by Step

For a rod moving perpendicular to BB:

ε=BLv=0.5×0.5×4=1 V\varepsilon = BLv = 0.5 \times 0.5 \times 4 = 1 \text{ V}

Total circuit resistance: rod 0.5Ω0.5 \Omega + external 2Ω=2.5Ω2 \Omega = 2.5 \Omega.

I=εRtotal=12.5=0.4 AI = \frac{\varepsilon}{R_{\text{total}}} = \frac{1}{2.5} = 0.4 \text{ A}

The current-carrying rod in field BB experiences a force Fmag=BILF_{\text{mag}} = BIL opposing its motion (Lenz’s law):

Fmag=0.5×0.4×0.5=0.1 NF_{\text{mag}} = 0.5 \times 0.4 \times 0.5 = 0.1 \text{ N}

To keep the rod at constant velocity, the applied force must balance this:

Fapplied=0.1 NF_{\text{applied}} = 0.1 \text{ N}

Mechanical power input: Pmech=Fapplied⋅v=0.1×4=0.4 WP_{\text{mech}} = F_{\text{applied}} \cdot v = 0.1 \times 4 = 0.4 \text{ W}.

Electrical power dissipated: Pelec=I2Rtotal=(0.4)2×2.5=0.4 WP_{\text{elec}} = I^2 R_{\text{total}} = (0.4)^2 \times 2.5 = 0.4 \text{ W}.

These match exactly — no surprise, since at constant velocity all the mechanical work goes into electrical heat.

Why This Works

The motional EMF ε=BLv\varepsilon = BLv comes from the magnetic Lorentz force on free charges inside the moving rod. Once current flows, that same field exerts an opposing force on the current-carrying rod — that’s Lenz’s law in action, ensuring energy conservation.

If the applied force were larger than 0.1 N0.1 \text{ N}, the rod would accelerate; smaller, it would decelerate. At exactly 0.1 N0.1 \text{ N} we get steady state, and the power identity Fv=I2RFv = I^2R is automatic.

Alternative Method

You can derive everything from flux. Flux through the circuit: Φ=BLx\Phi = BLx where xx is the rod’s position. Rate of change: dΦ/dt=BL(dx/dt)=BLvd\Phi/dt = BL(dx/dt) = BLv, which gives the EMF directly. Both methods are valid; “motional EMF = BLvBLv” is just the special case of Faraday’s law for a translating rod.

Energy conservation gives a useful shortcut: in any rod-on-rails problem at constant velocity, applied power = dissipated power. So Fapplied=I2R/vF_{\text{applied}} = I^2R/v. Worth remembering for MCQ-style questions where you just need the force.

Common Mistake

Students forget the rod’s own resistance and compute I=ε/RextI = \varepsilon/R_{\text{ext}} only — getting I=0.5 AI = 0.5 \text{ A} instead of 0.4 A0.4 \text{ A}. The rod is part of the circuit; its resistance always adds in series with the external load. Read the problem carefully — when the rod resistance is given, it must be included.

The other common error is sign confusion in Lenz’s law. The induced current creates a force opposing the rod’s motion. Many students draw the force in the direction of motion, then conclude no external force is needed — leading to absurd “free energy” answers.

Final answer: ε=1 V\varepsilon = 1 \text{ V}, I=0.4 AI = 0.4 \text{ A}, F=0.1 NF = 0.1 \text{ N}, P=0.4 WP = 0.4 \text{ W}.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next

AC generator — derive EMF equation ε = NBA ω sin(ωt)

medium

Displacement current — Maxwell's correction to Ampere's law

hard

Electromagnetic Induction: Diagram-Based Questions (7)

easy

Electromagnetic induction — eddy currents, applications and prevention

easy

Electromagnetic Induction: Edge Cases and Subtle Traps (5)

medium