A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5-0 m - s at right angles to the horizontal component of earth-s magnetic field 0-3 - 10-4 tesla- The emf induced across the ends of wire isA- 1-5 mV with higher potential at eastern endB- 1-5 mV with higher potential at western endC- 0-75 mV with higher potential at eastern endD- Zero

E = B_H lu = 0.3 × 10^ 4×10 ×5 = 15 × 10^ 4 = 1.5 × 10^ 3 v = 1.5 mV with higher potential at eastern end ...

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Standard XII

Physics

Motional EMF

A horizontal ...

Question

A horizontal straight wire 10m long extending from east to west is falling with a speed of 5.0 m/s at right angles to the horizontal component of earth's magnetic field $0.3 \times 10^{- 4}$ tesla. The emf induced across the ends of wire is

1.5 mV with higher potential at eastern end

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1.5 mV with higher potential at western end

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0.75 mV with higher potential at eastern end

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Zero

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Solution

The correct option is A 1.5 mV with higher potential at eastern end
$e = B_{H} l u$
$= 0.3 \times 10^{- 4} \times 10 \times 5$
$= 15 \times 10^{- 4}$
$= 1.5 \times 10^{- 3} v$
= 1.5 mV with higher potential at eastern end

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A horizontal straight wire 10m long extending from east to west is falling with a speed of 5.0 m/s at right angles to the horizontal component of earth's magnetic field 0.3×10−4 tesla. The emf induced across the ends of wire is

The correct option is A 1.5 mV with higher potential at eastern ende=BHlu =0.3×10−4×10×5 =15×10−4 =1.5×10−3v = 1.5 mV with higher potential at eastern end

A horizontal straight wire 10m long extending from east to west is falling with a speed of 5.0 m/s at right angles to the horizontal component of earth's magnetic field $0.3 \times 10^{- 4}$ tesla. The emf induced across the ends of wire is

The correct option is A 1.5 mV with higher potential at eastern end
$e = B_{H} l u$
$= 0.3 \times 10^{- 4} \times 10 \times 5$
$= 15 \times 10^{- 4}$
$= 1.5 \times 10^{- 3} v$
= 1.5 mV with higher potential at eastern end