Answer to Question #182688 in Electricity and Magnetism for Andrei

Question

1.   7.55-μC charged particle with a speed of 30.50 m/s is found in a uniform magnetic field with magnitude 1.2 T. Solve for the magnitude of the magnetic force exerted on the charged particle if the particle is moving perpendicular to the field.

 

Answer 1

To be given in question

q$=7.55\mu c$

$B=1.2Tesla$

$v=30.50meter/sec$

$\theta=90°$

To be asked in question

Magnetic force (F_B)=?

We know that

Magnetic force $F_{B}=q(v\times B)$

$F_{B}=qvBsin\theta$

$sin90°=1$

$F_{B}=7.55\times10^{-6}\times30.50\times1.2$

F_B=$2.7633\times10^{-04} N$