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A magnetic field is confined to a cylindrical region of radius R and changes with time as B = (2.00 t³ - 4.00 t² + 0.8) T. The geometry has r2 = 2R = 5.0 cm (so R = 2.5 cm) and a point P1 at r1 = 0.02 m inside the field region. Using the induced electric field, what is the magnitude of the force on an electron located at P2 (radius r2) at t = 2.00 s? (Take electron charge magnitude e = 1.6 x 10⁻¹⁹ C.)

1. 8.0 x 10⁻²¹ N
2. 1.6 x 10⁻²⁰ N
3. 4.0 x 10⁻²¹ N
4. 2.4 x 10⁻²⁰ N

Correct answer: 8.0 x 10⁻²¹ N

Solution

The changing flux induces a circulating electric field. For a loop of radius r2 = 5.0 cm, which is OUTSIDE the field region (R = 2.5 cm), the enclosed flux is B*pi*R², so E*(2 pi r2) = pi R² |dB/dt|, giving E = (R²/(2 r2))|dB/dt|. With dB/dt = 6t² - 8t, at t = 2.00 s: dB/dt = 6(4) - 8(2) = 24 - 16 = 8.0 T/s. Then E = (0.025² / (2*0.05)) * 8.0 = (6.25e-4/0.10)*8.0 = 6.25e-3 * 8.0 = 0.050 V/m. Force on electron F = eE = 1.6e-19 * 0.050 = 8.0 x 10⁻²¹ N. (For part b, at P1 inside: E = (r1/2)|dB/dt|; at t = 3.00 s, dB/dt = 6(9)-8(3)=30 T/s, so E = (0.02/2)*30 = 0.30 V/m, directed tangentially/circularly around the axis.)

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