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Tcecarenko [31]
3 years ago
5

A long straight wire carries current to the right of the page. A rectangular loop is positioned directly under the wire in the p

lane of the page and carries current in the clockwise direction.
The net force exerted by this loop on the straight wire with the current is directed:

a) towards the loop.
b) away from the loop.
c) to the right of the page.
d) to the left of the page.
e) is zero.
Physics
1 answer:
slavikrds [6]3 years ago
8 0

Answer:

Correct option A.

The net force exerted by this loop on the straight wire with the current is directed TOWARDS THE LOOP

Explanation:

The magnetic field exerts a force on a current-carrying wire in a direction given by the right-hand rule 1 (the same direction as that on the individual moving charges). This force can easily be large enough to move the wire since typical currents consist of very large numbers of moving charges.

Given that,

The wire's current is directed towards the right of the page.

The rectangular loops carry current in a clockwise direction.

Since the 'dot' field is increasing hence the induced magnetic field is 'cross', i.e. into the page and by the right-hand rule, the induced current is clockwise.

Then the magnetic field is into the page.

Since was known that

F= iL×B

Note the current is through the wire. Then, the length is in direction of the current.

Note: this equation gives the magnetic force that acts on a length L of a straight wire carrying a current (i) and immersed in a uniform magnetic field (B), that is perpendicular to the wire.

So, the magnetic field is always perpendicular to the current.

So using right hand rule,

F = i(L×B)

The length is to the right i.e. +x direction and the Magnetic field is perpendicular to the plane, i.e. in the +z direction

F = i (L•i × B•k)

F = iLB (i×k)

F = iLB•(-j)

F = -iLB•j

Then, the force is in the negative y-direction i.e. towards the loop.

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Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.3
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Answer:

844°C

Explanation:

The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.

We know that Fick's law is given by,

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Where \frac{\Delta c}{\Delta x} is the concentration of gradient

D is the diffusivity coefficient

and J is the flux of atoms.

In the other hand we have, that

D= D_0 e^{\frac{E_d}{RT}}

Where D_0 is the proportionality constant,

R is the gas constant, T the temperature and E_d is the activation energy.

Replacing the value of diffusivity coefficient in Fick's law we have,

J = -D_0 ^{\frac{E_d}{RT}}\frac{\Delta c}{\Delta x}

Rearrange the equation to get the value of temperature,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

We have all the values in our equation.

\Delta c = 0.664-0.339 = 0.325 C. cm^{-1}

\Delta x = 9.7*10^{-3}m

E_d = 82000J

D_0 = 6.5*10^{-7}m^2/s

J = 3.2*10^{-9}m^2/s

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Substituting,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

T=-\frac{-82000}{(8.31)ln(\frac{3.2*10^{-9}(9.7*10^{-3})}{6.5*10^{-7} (0.325)})}

T=1118.07K=844\°C

4 0
4 years ago
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