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Pavlova-9 [17]
3 years ago
12

A straight wire segment 5 m long makes an angle of 30° with a uniform magnetic field of 0.37 T. Find the magnitude of the force

on the wire if it carries a current of 2.9 A.
Physics
1 answer:
SIZIF [17.4K]3 years ago
7 0

Answer:

The magnitude of the force on the wire is 2.68 N.

Explanation:

Given that,

Length of the wire, L = 5 m

Magnetic field, B = 0.37 T

Angle between wire and the magnetic field, \theta=30^{\circ}

Current in the wire, I = 2.9 A

We need to find the magnitude of the force on the wire. The magnetic force in the wire is given by :

F=BIL\ \sin\theta\\\\F=0.37\ T\times 2.9\ A\times 5\ m\times \ \sin(30)\\\\F=2.68\ N

So, the magnitude of the force on the wire is 2.68 N. Hence, this is the required solution.

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The question is incomplete. I can help you by adding the information missing. They want you to calculate a) the radius of the cyclotron orbit for an electron with speed 1.0 * 10^6 m/s^2 and b) the radius of a cyclotron orbit for a proton with speed 5.0 * 10^4 m/s.

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Centripetal force, F = m*Ac = m * v^2 / R

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