Question

acts on
& A force of
1.732 x 10-2 N
a particle of
charge q moving with a velocity 1/1000th of the
velocity of light in a magnetic field of induction
2/
$\sqrt{3 }$
3 x 10-5 woblm? and perpendicular to the field
the 9 is ?​

Answer 1

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Concept:

• Magnetic fields

Given:

• F = 1.732×10^−2 N
• charge = q
• velocity = 1/1000ℎ of the velocity of light = 3*10^5 m/s
• Magnetic field = 2/√3*10^-5 Wb/m^2
• Angle θ = 90°

Find:

• The value of the charge q

Solution:

F = qvB sin θ

1.732×10^−2 = q3*10^5  * 2/√3*10^-5 sin 90°

q =√3* 1.732* 10^−2 /3*10^5 *10^-5

q = 5000 microcoloumbs

The value of q = 5000 microcoulombs.

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A force of 1.732×10^−2 N acts on a particle of charge q moving with a velocity 1/1000ℎ of the velocity of light in a magnetic field of induction      2/√3*10^-5 Wb/m^2 and perpendicular to the field. Then q is _______ .

Answer 2

Answer:

The answer is = 5000µc

The force acting on a moving charge in a magnetic field is determined by the linear combination of the charge's magnitude and the magnetic field's strength.

Explanation:

A charged particle travelling at speed v through an electric field E and a magnetic field B experiences the Lorentz force. After the Dutch physicist Hendrik A. Lorentz, the overall electromagnetic force F acting on the charged particle is known as the Lorentz force and is denoted by the equation F = qE + qv B.

The Lorentz force is the overall force exerted on a charge (q) travelling with velocity (v) in the presence of magnetic and electric fields (B and E, respectively).

$F = 1.732*10^{-2} N\\charge = q$

$velocity = 1/1000h$  

velocity of light $= 3*10^5 m/s$

Magnetic field $=\frac{2}{\sqrt{3*10^{-5} } } wb/m^{2}$

Angle θ $= 90$°

$F = 1.732*10^{-2} N$

$V=\frac{x}{1000^{th} } *10^{8} m/s$

   $=3*10^{5} m/s$

$F=qvBsin$θ

$q=F/VBsin$θ

θ $=90$°

$sin90$° $=1$

$q=\frac{1.732*10^{-2}N }{3*10^{5} *\frac{2}{\sqrt{3} }*10^{-5}sin90 }$

$q=0.5*10^{-2} C$

$q=5000$µc

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