Question

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. segment bc is an arc of a circle with radius 30.0 cm, and point p is at the center of cur- vature of the arc. segment dais an arc of a circle with radius 20.0 cm, and point p is at its cen- ter of curvature. segments cd and ab are straight lines of length 10.0 cm each

Answer 1

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 $r_{bc} = 30.0 \ cm =\frac{30}{100} = 0.3m$

        The radius  of arc da is $r_{da} = 20.0 \ cm = \frac{20}{100} = 0.20 \ m$

        The length of segment cd and ab is = $l = 10cm = \frac{10}{100} = 0.10 m$

The objective of the solution is to obtain the magnetic field

    Generally magnetic due to the current flowing in the arc is mathematically represented as

             $B = \frac{\mu_o I}{4 \pi r}$

 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  $\theta = 120^o = 120^o * \frac{\pi}{180} = \frac{2\pi}{3} rad$

     Now substituting values into formula for magnetic field for da

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

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

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

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