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Simran Bhatia Grade: 11
        A 20 cm long conducting rod is set into pure translation with a uniform velocity of 10 cm/s perpendicular to its length. A uniform magnetic field of magnitude 0.10 T exists in a direction perpendicular to the plane of motion. (a) Find the average magnetic force on the free electrons of the rod. (b) For what electric field inside the rod, the electric force on a free electron will balance the magnetic force? How is this electric field created? (c) Find the motional emf between the ends of the rod.
3 years ago

Answers : (1)

Navjyot Kalra
askIITians Faculty
654 Points
										Sol. l = 20 cm = 0.2 m

v = 10 cm/s = 0.1 m/s
B = 0.10 T
a) F = q v B = 1.6 × 10^–19 × 1 × 10^–1 × 1 × 10^–1 = 1.6 × 10^–21 N
b) qE = qvB
⇒ E = 1 × 10^–1 × 1 × 10^–1 = 1 × 10^–2 V/m
This is created due to the induced emf.
c) Motional emf = Bvℓ
= 0.1 × 0.1 × 0.2 = 2 × 10^–3 V
3 years ago
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