Question

A horizontal wire 0.1 m long carries a current of 5A from left to right. Find the magnitude and direction of the magnetic field, which support the weight of the wire. Assume wore to be of mass 3 ×10^-3 Kg m^-1.​

Answer 1

Answer:-

$\pink{\bigstar}$ The magnitude of the magnetic field is $\large\leadsto\boxed{\tt\blue{5.88 \times 10^{-3} \: T}}$

$\pink{\bigstar}$ The direction of the magnetic field is horizontally perpendicular to the wire.

• Given:-

• Horizontal wire of 0.1 m length carries current of 5A from left to right.

• Mass of the wire = 3 × 10-³ kg m-¹ .

• To Find:-

• The magnitude and direction of the magnetic field which supports the weight of the wire.

• Solution:-

Here,

• $\sf I = 5 A$

• $\sf l = 0.1 \: m$

• $\sf m = 3 \times 10^{-3} kg.m^{-1}$

The weight of the wire will be:-

➪ $\sf mg = 3 \times 10^{-3} \times 0.1 \times 9.8$

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➪ $\sf mg = 0.3 \times 10^{-3} \times 9.8$

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➪ $\bf mg = 2.94 \times 10^{-3} \: N$

★ The direction of the magnetic field will be horizontally perpendicular to the wire.

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★ Figure:-

$\setlength{\unitlength}{2.4mm}\begin{picture}(10,10)\thicklines\put(0,0){\vector(0,2){10}}\put(0,0){\vector(0,-2){10}}\put(0,0){\line(2,0){10}}\put(0,0){\line(-2,0){10}}\put(-8,0){\vector(2,0){5}}\put(1,1){\large\bf \vec{B} }\put(2,8){\large\bf F (magnetic)}\put(-6,-2){\large\bf I}\put(1,-10){\large\bf mg}\end{picture}$

We know,

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\green{F = ILB}}}$

where,

• F = Magnetic force
• I = Current
• L = Length of wire
• B = Magnetic field

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In equilibrium,

• F = mg

Hence,

• Substituting in the Formula:-

➪ $\sf 2.94 \times 10^{-3} = 5 \times 0.1 \times B$

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➪ $\sf 2.94 \times 10^{-3} = 0.5 B$

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➪ $\sf B = \dfrac{2.94 \times 10^{-3}}{0.5}$

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★ $\large{\bf\pink{B = 5.88 \times 10^{-3} \: T}}$

Answer 2

Answer:

★ Magnetic field (B) $⇝$ $\boxed{\tt{\red{5.88 \times 10^{-3} \: T}}}$

★ The direction of the magnetic field is horizontally perpendicular to the wire.

$\:$

Given :-

$\sf{Here, I=5A;\ell =0.1m;}$

Mass per unit length of the wire

$\sf{m\:(for\:1\:meter)=3×10^{−3}kg}$

$\:$

Solution :-

Therefore, weight of the wire,

➪ $\sf{mg=3×10^{−3}×0.1×9.8}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:=2.94×10^{−3}N\:--(i)}$

$\:$

Let B be the strength of the magnetic field applied.

➪ $\sf{F=BI\ell=B×5×0.1=0.5B\:--(ii)}$

$\:$

In equilibrium, (using i and ii)

$\sf{➪\:F=mg}$

$\sf{➪\:0.5B=2.94×10^{−3}}$

$\sf{➪\:B= \dfrac{2.94×10^{−3}}{0.5}}$

$\:\:\:\:\:\:\:\:\:=$ $\bold{\red{ 5.88×10^{−3} T}}$