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lana [24]
2 years ago
9

A charge feels a 7.89 x 10-7 N force when it moves 2090 m/s at a 29.4° angle to a 4.23 x 10-3 T magnetic field. What is the amou

nt of the charge?
Physics
2 answers:
tamaranim1 [39]2 years ago
6 0

We have that the amount of the charge q is

q=1.8*10^{-7}

From the Question we are told that

Force F=7.89 x*10^{-7}

Velocity V=2090m/s

Angle \theta=29.4

Magnetic field B=4.23 * 10^{-3} T

Generally, the equation for Force F is mathematically given by

F=qVBsin\theta\\\\q=\frac{F}{VBsin\theta}

q=\frac{7.89 x*10^{-7}}{4.23 * 10^{-3} T*sin29.4*2090m/s}

q=1.8*10^{-7}

In conclusion

The amount of the charge q is

q=1.8*10^{-7}

For more information on this visit

brainly.com/question/1470439

sasho [114]2 years ago
3 0

From the question, the given charge moving within the magnetic field would have a value of 1.82 x 10^{-7} C.

A charge is either a positively or negatively charged particle. When a charge is moving through a magnetic field, it would experience a force which depends on the amount of the charge, its speed, the magnetic field strength and angle. The force on a charge in a magnetic field can be determined by:

F = qvBSin θ

where: q is the charge, v is its velocity/ speed in the field and θ is the angle of deflection of the charge.

Given that: F = 7.89 x 10^{-7} N, v = 2090 m/s, B = 4.23 x 10^{-3} T and θ = 29.4°.

Thus, the amount of the charge can be determined by;

q = \frac{F}{vBsin O}

  = \frac{7.89*10^{-7} }{2090*4.23*10^{-3}* Sin 29.4 }

  = \frac{7.89*10^{-7} }{4.340}

q = 1.818 x 10^{-7}

q = 1.82 x 10^{-7} C

Therefore, the amount of the charge that moved within the magnetic field is 1.82 x 10^{-7} C.

For more understanding, visit: brainly.com/question/22403676

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