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A square loop slides into a magnetic field, moves through it and slides out. One diagram tests flux, Faraday's law, Lenz's law and graph reading, and most marks are lost on the graphs. This post builds the flux, EMF, current and force graphs for one loop with numbers, then shows the variations examiners use.
Where is this in the syllabus?
In the NCERT Class 12 chapter Electromagnetic Induction, Example 6.4 asks for the current direction in loops moving into and out of a field, and Example 6.5(c) compares a rectangular and a circular loop leaving a field and concludes that the EMF is constant only for the rectangular loop, because only its area inside the field changes at a steady rate. Exercise 6.4 is a numerical version.
The JEE (Main) 2026 syllabus lists "Faraday's law, induced emf and current, Lenz's law", and the NEET (UG) 2026 syllabus has the same topics. Check the syllabus for your own exam year.
Why are induction graphs hard?
In a study of 895 high school and college students, Beichner (1994) found consistent difficulties with motion graphs, including reading a graph as a picture and confusing slope with height. Induction graphs set the same trap. The EMF is the slope of the flux graph, not its height. When the flux is largest, with the loop fully inside the field, the EMF is zero.
Worked example: the set-up
A square loop of side L = 0.20 m and resistance R = 2.0 Ω moves to the right at a steady v = 0.10 m/s. A uniform field B = 0.50 T points into the page and fills a region 0.50 m wide. The leading edge reaches the field at t = 0.
0 to 2 s, entering. The area inside the field is L × vt, so Φ = BLvt. Flux rises steadily to BL² = 0.50 × 0.040 = 0.020 Wb, reached when the loop has moved its own length, 0.20 m, at t = 2 s.
2 to 5 s, fully inside. Flux stays at 0.020 Wb until the leading edge reaches the far boundary, 0.50 m from the start, at t = 5 s.
5 to 7 s, leaving. Flux falls steadily to zero.
After 7 s. Flux is zero.
What are the EMF and current?
Only the edge that is cutting the boundary matters. ε = BLv = 0.50 × 0.20 × 0.10 = 0.010 V, or 10 mV. Check it as the slope: 0.020 Wb ÷ 2 s = 0.010 V.
The current is I = ε/R = 0.010 ÷ 2.0 = 5.0 mA.
For direction, use Lenz's law. While entering, flux into the page is increasing, so the induced current makes a field out of the page inside the loop: anticlockwise as you look at the page. While leaving, flux into the page is decreasing, so the current is clockwise.
Taking anticlockwise as positive, the EMF graph is +10 mV from 0 to 2 s, zero from 2 to 5 s and −10 mV from 5 to 7 s. Two flat pulses of opposite sign.
Which way does the magnetic force point?
While entering, the leading edge is the side whose force counts; the forces on the top and bottom sides cancel. F = BIL = 0.50 × 0.0050 × 0.20 = 5.0 × 10⁻⁴ N, pointing left, against the motion. While leaving, the current has reversed, but now it is the trailing edge that sits in the field, so the force still points left. The magnetic force always opposes the motion. To keep the speed steady, you must push with 5.0 × 10⁻⁴ N during both stages.
Two checks:
Power. Fv = 5.0 × 10⁻⁴ × 0.10 = 5.0 × 10⁻⁵ W, and I²R = (0.0050)² × 2.0 = 5.0 × 10⁻⁵ W. The work you do becomes heat. Over the two 2 s stages that is 2.0 × 10⁻⁴ J.
Charge. q = ΔΦ/R = 0.020 ÷ 2.0 = 0.010 C per stage, the same as 5.0 mA × 2 s. The charge does not depend on the speed.
What changes when the field region is narrower than the loop?
Keep the loop but make the field region only 0.10 m wide. The leading edge crosses it from 0 to 1 s, and the flux rises to 0.50 × 0.20 × 0.10 = 0.010 Wb. From 1 to 2 s the field strip sits wholly inside the loop and the flux is constant. From 2 to 3 s the trailing edge crosses and the flux falls. The EMF is still ±10 mV, because it depends only on the edge doing the cutting, but each pulse now lasts 1 s.
For non-rectangular loops, the cutting length changes. As NCERT's Example 6.5(c) says, a circular loop's area inside the field does not change at a steady rate, so its EMF varies as it leaves.
For the direction rules on their own, see our post on finding the induced current direction with Lenz's law. For a rod that speeds up or slows down, see motional EMF for a rod on rails.
Common mistakes
Drawing the EMF graph with the same shape as the flux graph.
Reversing the force when the loop leaves. The current reverses; the force does not.
Forgetting the zero-EMF stage while the loop is fully inside.
Forgetting to multiply by N for a coil of N turns.
If you find it hard to picture the flux changing as the loop moves, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
A practice routine
For every loop problem, sketch Φ against t before writing any formula, then sketch its slope.
Redo the worked example with a 0.10 m field region and check your graphs against the section above.
Solve NCERT Exercise 6.4. You should get 2.4 × 10⁻⁴ V lasting 2 s when the velocity is normal to the longer side, and 0.6 × 10⁻⁴ V lasting 8 s when it is normal to the shorter side.
Check answers with power (Fv = I²R) and charge (q = ΔΦ/R).
Tonight, draw the four graphs for the worked example from memory and compare.
