A straight, horizontal wire of mass $10$ mg and length $1.0$ m carries a current of $2.0$ A. What minimum magnetic field B?

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Solution

Given mass $M = 10 m g = 10^{- 5} k g$ length $l = 1 m$ and current $I = 2.0 A$
We know magnetic field $B = \frac{M g}{I l} = \frac{10^{- 5} \times 10}{2 \times 1} = 5 \times 10^{- 5}$

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A straight, horizontal wire of mass 10 mg and length 1.0 m carries a current of 2.0 A. What minimum magnetic field B?

Given mass M=10mg=10−5kg length l=1m and current I=2.0AWe know magnetic field B=MgIl=10−5×102×1=5×10−5

A straight, horizontal wire of mass $10$ mg and length $1.0$ m carries a current of $2.0$ A. What minimum magnetic field B?

Given mass $M = 10 m g = 10^{- 5} k g$ length $l = 1 m$ and current $I = 2.0 A$
We know magnetic field $B = \frac{M g}{I l} = \frac{10^{- 5} \times 10}{2 \times 1} = 5 \times 10^{- 5}$