Question

A point charge of 5.7 µc moves at 4.5 × 105 m/s in a magnetic field that has a field strength of 3.2 mt, as shown in the diagram. a field of circles labeled b. there is a positively charged particle with a velocity vector pointing 37 degrees north of east labeled v. what is the magnitude of the magnetic force acting on the charge? 6.6 × 10–3 n 4.9 × 10–3 n 4.9 × 103 n 6.6 × 103 n

Answer 1

The magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

What is magnetic force?

Magnetic force, attraction or repulsion that arises between electrically charged particles because of their motion.The magnitude of the magnetic force acting on the charge is given by:

$\rm F=qvBsin\theta$

where

The magnitude of the charge   $q=5.7\ \mu C =5.7\times 10^{-6} \ C$

The velocity of the charge        $v=4.5\times 10^5\ \frac{m}{s}$

The magnitude of the magnetic field   $3.2\ mT=0.0032\ T$

The angle between the directions of v and B  $\theta =90^o-37^o=53^o$

By substituting the values we will get:

$F=(5.7\times 10^}-6})(4.5\times 10^5)(0.0032)(Sin53^o)$

$F=6.6\times 10^{-3}\ N$

Hence the magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

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