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Juli2301 [7.4K]
2 years ago
9

Help me please in this

Physics
1 answer:
vfiekz [6]2 years ago
7 0
U=- 2.5ms-¹
g=9.8ms-²
t=3s
V=- u +gt
V= 2.5+(9.8×3) = 2.5 + 29.4 = 31.9 ms-¹
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Two wooden boxes of equal mass but different density are held beneath the surface of a large container of water. Box A has a sma
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Calculate the mag-netic field (magnitude and direc-tion) at a point p due to a current i=12.0 a in the wire shown in fig. p28.68
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Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The magnitude is   B= 4.2 *10^ {-6}T , the direction is into the page

Explanation:

From the question we are told that

        The current  is i = 12.0 A

        The radius of arc  bc is r_{bc} = 30.0 \ cm =\frac{30}{100} = 0.3m

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 Here I is the current

         \mu_o is the permeability of free space with a value of 4\pi *10^{-7}T \cdot m/A

            r is the distance

Considering Arc da

         B_{da} = \frac{\mu_o I}{4 \pi r_{da}} \theta

Where \theta is the angle the arc da makes with the center  from the diagram its value is  \theta = 120^o = 120^o * \frac{\pi}{180} = \frac{2\pi}{3} rad

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                    B_{da} = \frac{4\p *10^{-7} * 12}{4 \pi (0.20)}[\frac{2 \pi}{3} ]

                           = \frac{10^{-7} * 12}{0.20} * [\frac{2 \pi}{3} ]

                   B_{da}= 12.56*10^{-6} T

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to da is into the pages of the paper

Considering Arc bc

             B_{bc} = \frac{\mu_o I}{4 \pi r_{bc}} \theta

Substituting value

          B_{bc} = \frac{4 \pi *10^{-7} * 12}{4 \pi (0.30)} [\frac{2 \pi}{3} ]

                B_{bc}= 8.37*10^{-6}T

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to bc is out of  the pages of the paper

Since the line joining P to segment bc and da makes angle = 0°

     Then the net magnetic field would be

                 B = B_{da} - B{bc}

                     = 12.56*10^{-6} - 8.37*10^{-6}

                     = 4.2 *10^ {-6}T

       Since B_{da} > B_{bc} then the direction of the net charge would be into the page

 

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