Question
Asked Aug 1, 2019
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An electron moves through a uniform magnetic field given by B vector= Bx i+(4.26 Bx)j. At a particular instant, the electron has velocity v vector = (2.12i+4.91j) m/s and the magnetic force acting on it is (7.80 × 10-19 )k N. Find Bx.

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Expert Answer

Step 1

Given

The magnetic field B = Bx i + (4.26 Bx) j

The velocity of the electron v = (2.12 i + 4.91 j) m/s

The force acting on the electron F = (7.80 x 10-19 k ) N

Known

The magnitude of the charge of electron q = e = 1.6 x 10-19 C

 

Step 2

Understanding

The electron moves in the xy plane

The magnetic field is acting in the xy plane

The force is acting in the positive z axis direction

 

Formula Used

 

When a charges particle is moving in a magnetic field, the charged particle will experience a force in the direction perpendicular to both the velocity and the magnetic field. This force is called magnetic Lorentz force

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F -xB)

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Step 3

Solution

 

...
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F -xB) F fF B) i j 2.12 4.91 0 4.26B 0 B 7.80 x 1019 k [(2.12x 4.26B.)-(4.91*B -1.6x109 -4.875 4.1212B_k B-.187

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Magnetism

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