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CaHeK987 [17]
3 years ago
11

Joe and Matt fight over the last piece of sushi. Joe uses his chopsticks to

Physics
1 answer:
dimaraw [331]3 years ago
4 0

Answer:

Matt

Explanation:

because he has more force

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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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since current density is given

J=kr

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di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr

integrating on both sides between the appropriate limits,

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

Magnetic field can be found by using Ampere's law

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

for points inside the wire ( r<a)

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,

B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=&#10;\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}&#10;\\\\B_{in}=\frac{\mu_0kl^2}{3}&#10;

now using equation 1, putting the value of k,

B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}&#10;\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}&#10;

B)

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

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

applying the Ampere's law

\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;

again using,equaiton 1,

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}

8 0
3 years ago
A block of stone measures 15cm x 15cm x 20 cm.What is the total surface area of the stone
Sidana [21]

This should help look at the pictures?

7 0
4 years ago
If the ball shown in the figure lands in 1.0 s, about what height was it thrown from? ​
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Answer:

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6 0
2 years ago
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In a rigid container, as the temperature of a gas decreases, the pressure of the gas will decrease. t f
Sergio039 [100]
The answer is true: the pressure of a gas will decrease as temperature decreases in a rigid container.

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A 52.5-turn circular coil of radius 5.35 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0
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Answer:

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6 0
3 years ago
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