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Eddi Din [679]
3 years ago
6

A charge of q= +15 uC moves in a Northeast direction with a speed 5 m/s, 25 degrees East of a magnetic field, pointing North, wi

th a magnitude of 0.08 T. Find the force (magnitude and direction) on the charge.
Physics
1 answer:
grin007 [14]3 years ago
6 0

Explanation:

It is given that,

Magnitude of charge, q=15\ \mu C=15\times 10^{-6}\ C

It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.

Magnetic field, B=0.08\ j

Velocity, v=(5\ cos25)i+(5\ sin25)j

v=[(4.53)i+(2.11)j]\ m/s

We need to find the magnitude of force on the charge. Magnetic force is given by :

F=q(v\times B)

F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]

<em>Since</em>, i\times j=k\ and\ j\times j=0

F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]

F=0.00000543\ kN

F=5.43\times 10^{-6}\ kN

So, the force acting on the charge is 5.43\times 10^{-6}\ kN and is moving in positive z axis. Hence, this is the required solution.

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=======

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We apply the Law of Ohm:

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<u></u>

.

<u></u>

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