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In a 1.35 T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50 μC and initially moving northward at 4.68 km/s is deflected toward the east.

a. What is the sign of the charge of this particle? Make a sketch to illustrate how you found your answer.
b. Find the magnetic force on the particle.

Respuesta :

Answer:

a

The sign on the charge is positive sign

According to Flemmings Left Hand rule

     The index finger which represents the Magnetic field(From the question ) is pointing vertically upward so the middle finger would be pointing east ward i.e in the positive x-axis and the  thumb also would be pointing toward east and it represents the force acting on the charge

    The diagram that illustrates this is shown in the first uploaded image

b

The magnetic force is given as  

       [tex]F = 0.053 \ N[/tex]

Explanation:

From the question we are told that

         The magnitude of the magnetic field is [tex]B = 1.35 T[/tex]

         The  magnitude of the charge is  [tex]q= 8.50 \mu C = 8.50*10^ {-6}C[/tex]

          The speed is  [tex]v= 4.68 km/s[/tex]

Generally the force is mathematically represented as

            [tex]F = q * v* B[/tex]

Substituting values

              [tex]= 8.5*10^{-6}* 1.35 *4.68 *1000[/tex]

              [tex]= 0.053 N[/tex]

Ver imagen okpalawalter8