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प्रश्न
Choose the correct option.
A conducting thick copper rod of length 1 m carries a current of 15A and is located on the Earth's equator. There the magnetic flux lines of Earth's magnetic field are horizontal, with the field of 1.3 x 10-4T, south to north. The magnitude and direction of the force on the rod, when it is oriented so that current flows from west to east are ______.
पर्याय
14 × 10-4N, downward
20 × 10-4N, downward
14 × 10-4N, upward
20 × 10-4N, upward
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उत्तर
A conducting thick copper rod of length 1 m carries a current of 15 A and is located on the Earth's equator. There the magnetic flux lines of Earth's magnetic field are horizontal, with the field of 1.3 × 10-4T, south to north. The magnitude and direction of the force on the rod, when it is oriented so that current flows from west to east are 20 × 10-4N, upward.
Explanation:
Express the relation for the force acting on a current-carrying wire due to a perpendicular magnetic field.
`vecF = i * L vec× B`
Here, F is force, B is the perpendicular magnetic field, i is current and L is the length of the conductor.
Substitute 1.3 × 10−4T for B, 1 m for L and 15A for I in the equation to find the magnitude of the force.
F = `(1.3 xx 10^-4T) xx (15A) xx (1m)`
F = `19.5 xx 10^-4N`
F ≈ `20 xx 10^-4N`
The direction of the force is found by using Fleming’s left-hand rule where the current corresponds to the middle finger, the field corresponds to the pointing finger and the force corresponds to the thumb. Hence the force is upwards.
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