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KengaRu [80]
4 years ago
9

What to do with a brand new mobile phone battery

Physics
1 answer:
Rashid [163]4 years ago
7 0
Place it in the back of your phone where the old battery was

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A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const
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Answer:

Explanation:

we can consider an element of radius r < a and thickness dr.  and Area of this element is

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\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr&#10;\\\\&#10;I=\frac{2\pi\,ka^3}{3} -------------------------------(1)

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\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}

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now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}&#10;

by symmetry \vec{B} will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

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B)

now, for points outside the wire ( r>a)

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\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}&#10;

by symmetry \vec{B} will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)&#10;\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}&#10;\\\\B_{out}=\frac{\mu_0ka^3}{3r}&#10;

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B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}&#10;\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}

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4 years ago
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Answer:

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You're welcome if I helped!

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