<p>Review Conceptual Example 2 as an aid in understanding this problem. The draw

Question

<p>Review Conceptual Example 2 as an aid in understanding this problem. The drawing shows a positively charged particle entering a 0.72-T magnetic field. The particle has a speed of 250 m/s and moves perpendicular to the magnetic field. Just as the particle enters the magnetic field, an electric field is turned on. What must be the magnitude of the electric field such that the net force on the particle is twice the magnetic force?&#160;<br /><br />&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;.&#160;&#160; .&#160;&#160;&#160; .&#160;&#160; .&#160;&#160; .-b<br />q------&#160;&gt;.&#160;&#160; .&#160;&#160;&#160; .&#160;&#160; .&#160;&#160; .&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;</p>
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Explanation / Answer

      The magnetic field, B = 0.72 T       The speed of the particle, v = 250 m/s       The angle between speed v and magnetic field B, = 90o ______________________________________________________       The electrostatic force related with electric field is                            FE= qE       The magnetic force related with magnetic field is                           FB= qvBsin _______________________________________________________      When the net force on the particle is twice the magnetic force,       then the magnetic force is equal to magnitude of the electric force.       Such that, FE= FB       So, from above equations                       qE = qvBsin       Therefore, the magnetude of the electric field is                      E = vBsin90o                          = (250 m/s)(0.72 T)(1)                          = 180 V/m       Direction: In this case, according to right hand rule                         the direction of the electric field is pointed to the                        bottom of the page. _______________________________________________________ Note: The other units for electric field is N/C                                 = (250 m/s)(0.72 T)(1)                          = 180 V/m       Direction: In this case, according to right hand rule                         the direction of the electric field is pointed to the                        bottom of the page. _______________________________________________________ Note: The other units for electric field is N/C              

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