Question

the wire is 5.20 cm from the charge and carries a current of 68.5 A in a direction opposite to that of the moving charge. Calculate the magnitude of the force on the charge.

Answer 1

Complete Question

A 6.00-µc charge is moving with a speed of 7.50x10^4 m/s parallel to a very long, straight wire. the wire is 5.20 cm from the charge and carries a current of 68.5 A in a direction opposite to that of the moving charge. Calculate the magnitude of the force on the charge.

Answer:

The value is $F =1.20 *10^{-4} \ N$

Explanation:

From the question we are told that

     The magnitude of the charge is  $Q = 6.00\muC = 6.00 *10^{-6} \ C$

      The speed is  $v = 7.50 *10^{4} \ m/s$

      The distance of the wire from the charge is  $d = 5.20 \ cm = 0.0520 \ m$

       The current flowing through the wire in the opposite direction to the charge  is  $I _2 = 68.5 \ A$

  Gnerally the magnitude of force on that charge is mathematically represented as

      $F = qvB$

Here B is magnetic field which is mathematically represented as

              $B = \frac{\mu_o * I}{2\pi d}$

Here  $\mu_o$ is the permeability of free space with value  $\mu_o = 4\pi * 10^{-7} N/A^2$

 So

       $F = qv[\frac{\mu_o * I}{2\pi d}]$

=>     $F = 6.0*10^{-6} * (7.50 *10^{4})[\frac{4\pi * 10^{-7} * 68.5 }{2* 3.142 * 0.0520 }]$      

=>    $F =1.20 *10^{-4} \ N$