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My name is Ann [436]
3 years ago
10

A horizontal wire of length 0.53 m. carrying a current of 7.5 A. is placed in a uniform external magnetic field. When the wire i

s horizontal, it experiences no magnetic force. When the wire is tilted upward at an angle of 19°, it experiences a magnetic field of 4.4x10^-3N. Determine the magnitude of the external magnetic field.
Physics
1 answer:
Lady_Fox [76]3 years ago
6 0

Answer:

3.4 mT

Explanation:

L = 0.53 m

i = 7.5 A

Theta = 19 degree

F = 4.4 × 10^-3 N

Let B be the strength of magnetic field.

Force on a current carrying conductor placed in a magnetic field.

F = i × L × B × Sin theta

4.4 × 10^-3 = 7.5 × 0.53 × B × Sin 19

B = 3.4 × 10^-3 Tesla

B = 3.4 mT

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<h2>82.353 km/hr</h2>

Explanation:

       The driver travels 135 km towards East in 1.5 hr. He stops for 45 min. He again travels 215 km towards East in 2.0 hr.

       The total displacement of the driver in the given time is ths sum of individual displacements, because all the displacements are in the same directon.

       Total displacement = 135\text{ }km\text{ }East\text{ }+\text{ }0\text{ }km\text{ }+\text{ }215\text{ }km\text{ }East\text{ }=\text{ }350\text{ }km\text{ }East

       Total time travelled = 1.5\text{ }hr\text{ }+\text{ }45\text{ }min\text{ }+\text{ }2.0\text{ }hr\text{ }=\text{ }90\text{ }min\text{ }+\text{ }45\text{ }min\text{ }+\text{ }120\text{ }min\text{ }=\text{ }255\text{ }min\text{ }=\text{ }4\text{ }hr\text{ }15\text{ }min\text{ }=\text{ }4.25\text{ }hr

      \text{Average velocity = }\dfrac{\text{Total displacement}}{\text{Total time taken}}=\dfrac{350\text{ }km}{4.25\text{ }hr}=82.353\text{ }\frac{km}{hr}

∴ Driver's average velocity = 82.353\text{ }\frac{km}{hr}

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If a diffraction grating has 3700 lines per cm, what is the spacing d between lines
Sonja [21]
So first we find the gap between the slits by the formula d=1/N 

<span>N is number of lines per metre so 3700 line/cm = 370000 lines/m </span>
<span>So d=2.7*10^-6 </span>

<span>Now we use the formula dsin(angle)=n(wavelength) </span>

<span>d is the same </span>
<span>n is the order of the diffraction pattern </span>

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<span>=[(2.7*10^-6)*sin30]/3 </span>
<span>=4.5*10^-7 m</span>
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F_{engine}=425+840=1265 N

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