Question

A straight wire of mass 0.5
kg and length 2 m carries a
current of l ampere. If it is suspended in mid air where
uniform magnetic field of it exist perpendicular to length of wire, then value of I will be (g = 10 m/s?)
5 A
2.5 A
2 A
0.5 A​

Answer 1

Answer:

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Explanation:

Mass of wire (m) = 200 grams or 0.2 kg

Mass of wire (m) = 200 grams or 0.2 kglength (L) = 1.5 m

Mass of wire (m) = 200 grams or 0.2 kglength (L) = 1.5 mCurrent (I) = 2 A

Mass of wire (m) = 200 grams or 0.2 kglength (L) = 1.5 mCurrent (I) = 2 AAcceleration due to gravity (g) = 9.8 m/s^2

Mass of wire (m) = 200 grams or 0.2 kglength (L) = 1.5 mCurrent (I) = 2 AAcceleration due to gravity (g) = 9.8 m/s^2The weight of the wire will be balanced by the magnetic force acting in the opposite direction.

Mass of wire (m) = 200 grams or 0.2 kglength (L) = 1.5 mCurrent (I) = 2 AAcceleration due to gravity (g) = 9.8 m/s^2The weight of the wire will be balanced by the magnetic force acting in the opposite direction.Let the magnetic field be B.

Mass of wire (m) = 200 grams or 0.2 kglength (L) = 1.5 mCurrent (I) = 2 AAcceleration due to gravity (g) = 9.8 m/s^2The weight of the wire will be balanced by the magnetic force acting in the opposite direction.Let the magnetic field be B.mg = ILB

Mass of wire (m) = 200 grams or 0.2 kglength (L) = 1.5 mCurrent (I) = 2 AAcceleration due to gravity (g) = 9.8 m/s^2The weight of the wire will be balanced by the magnetic force acting in the opposite direction.Let the magnetic field be B.mg = ILBB = \frac{mg}{IL}

Mass of wire (m) = 200 grams or 0.2 kglength (L) = 1.5 mCurrent (I) = 2 AAcceleration due to gravity (g) = 9.8 m/s^2The weight of the wire will be balanced by the magnetic force acting in the opposite direction.Let the magnetic field be B.mg = ILBB = \frac{mg}{IL}B = \frac{0.2 \times 9.8}{2 \times 1.5}

Mass of wire (m) = 200 grams or 0.2 kglength (L) = 1.5 mCurrent (I) = 2 AAcceleration due to gravity (g) = 9.8 m/s^2The weight of the wire will be balanced by the magnetic force acting in the opposite direction.Let the magnetic field be B.mg = ILBB = \frac{mg}{IL}B = \frac{0.2 \times 9.8}{2 \times 1.5}B = 0.653 T

Mass of wire (m) = 200 grams or 0.2 kglength (L) = 1.5 mCurrent (I) = 2 AAcceleration due to gravity (g) = 9.8 m/s^2The weight of the wire will be balanced by the magnetic force acting in the opposite direction.Let the magnetic field be B.mg = ILBB = \frac{mg}{IL}B = \frac{0.2 \times 9.8}{2 \times 1.5}B = 0.653 TThe value of uniform magnetic field is 0.653 T.

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