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Airida [17]
3 years ago
6

A current-carrying wire 0.50 m long is positioned perpendicular to a uniform magnetic field. If the current is 10.0 A and there

is a resultant force of 3.0 N on the wire due to the interaction of the current and field, what is the magnetic field strength?
Physics
1 answer:
____ [38]3 years ago
7 0

<em>Answer</em>


0.6 teslas


<em>Explanation</em>

When a conductor is inside a magnetic field it experiences a force given by;

Force = ILBsinθ


Where I⇒ current

L ⇒length of the conductor

B ⇒ magnetic field strength

θ ⇒ Angle between the conductor and magnetic field.


F = ILBsinθ


When θ = 90°, Then sin 90 =1 and the formula becomes;

F =ILB

3 = 10 × 0.5 × B

3 = 5B

B = 3/5

= 0.6


magnetic field strength = 0.6 teslas


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A 13.0-g wad of sticky clay is hurled horizontally at a 110-g wooden block initially at rest on a horizontal surface. The clay s
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Answer:

v_{ic}=92.53 m/s

Explanation:

We need to apply conservation of momentum and energy to solve this problem.

<u>Conservation of momentum</u>

p_{i}=p_{f}

m_{c}v_{ic}=(m_{c}+m_{w})V (1)

  • m(c) is the mass of stick clay
  • m(w) is the mass of the wooden block
  • v(ic) is the initial velocity of clay
  • V is the final velocity of the system clay plus wood.

<u>Conservation of total energy</u>

The change in kinetic energy is equal to the change in internal energy, in our case it would be the energy loss due to the friction force. Let's recall the definition of work, it is the dot product between force and displacement, Therefore:

\Delta E=W

\frac{1}{2}(m_{c}+m_{w})V^{2}=F_{friction}*d

\frac{1}{2}(m_{c}+m_{w})V^{2}=\mu (m_{c}+m_{w})gd

We can find V from this equation:

V=\sqrt{2\mu gd}=\sqrt{2*0.65*9.81*7.5}=9.78 m/s

Now, let's put V into the equation (1) and find v(ic)

v_{ic}=\frac{(m_{c}+m_{w})V}{m_{c}}=\frac{123*9.78}{13}=92.53 m/s

I hope it helps you!  

<u />

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