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disa [49]
3 years ago
5

An electron travels at a speed of 16,748.76 m/s through a uniform magnetic field whose magnitude is 0.0177 T. What is the magnit

ude of the magnetic force on the electron if its velocity vector and the magnetic field vector are make an angle of 59.24° ?
Physics
1 answer:
MAVERICK [17]3 years ago
7 0

Answer:

Magnetic force, F=4.07\times 10^{-17}\ N

Explanation:

It is given that,

Speed of electron, v = 16748.76 m/s

Magnetic field, B = 0.0177 T

Angle between velocity vector and the magnetic field vector are make an angle of 59.24°. Magnetic force is given by :

F=qvB\ sin\theta

F=1.6\times 10^{-19}\times 16748.76\times 0.0177 \ sin(59.24)

F=4.07\times 10^{-17}\ N

So, the magnetic force on the electron is 4.07\times 10^{-17}\ N. Hence, this is the required solution.

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Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec
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A sound wave of 70 cm travels 840 m in 2.5 sec. What is the velocity and frequency of sound?
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5 0
3 years ago
11. एक समान चुम्बकीय क्षेत्र में लम्बवत प्रवेश करने वाले किसी आवेशित कण द्वारा प्राप्त वृत्तीय पथ की
mojhsa [17]

Answer:

<h2> r=mv/Be</h2>

Explanation:

If a positive charge enters a magnetic field at 90 degrees the charge is deflected in a circular path by a force that acts perpendicular to it in line with Flemings right-hand rule

to derive the radius of the path of the charge we apply

F= mv^2/r=Bev

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m= mass of the electronic charge

e=charge

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v=average speed

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rearranging we have

r=mv^2/Bev

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5 0
3 years ago
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