A point charge of 5. 0 Ă— 10â€""7 C moves to the right at 2. 6 Ă— 105 m/s in a magnetic field that is directed into the screen and has a field strength of 1. 8 Ă— 10â€""2 T. What is the magnitude of the magnetic force acting on the charge? 0 N 2. 3 Ă— 10â€""3 N 23 N 2. 3 Ă— 1011 N.

Respuesta :

The magnitude of the magnetic force acting on the charge which moves to the right is 0 N.

Given to us,

the charge [tex]q[/tex] =  [tex]5\times 10^{-7}[/tex] C,

the velocity [tex]v[/tex] = [tex]2.6\times 10^5[/tex]  m/sec,

the magnetic field [tex]B[/tex] = [tex]10^{-2}[/tex] T,

angle between the direction of v and B [tex]\theta[/tex] = 0,

Magnetic force is as important as the electrostatic or Coulomb force. The magnitude of the magnetic force F on a charge q moving at a velocity of v in a magnetic field of strength B is given by

[tex]\begin{aligned}F&=qvB\ sin\Theta\\&= 5\times10^{-7}\times2.6\times10^5\times10^{-2} \times sin(0)\\&= 0\ N\\\end{aligned}[/tex]

Hence, the magnitude of the magnetic force acting on the charge which moves to the right is 0 N

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