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Salsk061 [2.6K]
2 years ago
7

A point charge of 5.0 x 10^-7 C moves to the right at 2.6 x 10^5 m/s in a magnetic field that is directed into the screen and ha

s a field strength of 1.8 x 10^-2 T.
What is the magnitude of the magnetic force acting on the charge?
Physics
1 answer:
-BARSIC- [3]2 years ago
5 0

The magnitude of the magnetic force acting on the charge is 2.34×10⁻³ N.

<h3>What is magnetic force?</h3>

A magnetic force is the force that act in a magnetic field.

To calculate the magnetic force, we use the formula below.

Formula:

  • F = qvB.........Equation 1

Where:

  • F = magnetic force
  • q = point charge
  • v = Velocity of the the charge
  • B = Field strength

From the question,

Given:

  • q = 5.0×10⁻⁷ C
  • v = 2.6×10⁵ m/s
  • B = 1.8×10⁻² T

Substitute these values into equation 2

  • F = (5.0×10⁻⁷)(2.6×10⁵)(1.8×10⁻²)
  • F = 23.4×10⁻⁴
  • F = 2.34×10⁻³ N

Hence, the magnitude of the magnetic force acting on the charge is 2.34×10⁻³ N.

Learn more about magnetic force here: brainly.com/question/2279150

#SPJ12

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y = 0.628

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Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

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A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

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d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

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