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.
1 answer:
Explanation:
It is given that,
Magnitude of charge, 
It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.
Magnetic field, 
Velocity, 
![v=[(4.53)i+(2.11)j]\ m/s](https://tex.z-dn.net/?f=v%3D%5B%284.53%29i%2B%282.11%29j%5D%5C%20m%2Fs)
We need to find the magnitude of force on the charge. Magnetic force is given by :

![F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]](https://tex.z-dn.net/?f=F%3D15%5Ctimes%2010%5E%7B-6%7D%5B%284.53i%2B2.11j%29%5Ctimes%200.08%5C%20j%5D)
<em>Since</em>, 
![F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]](https://tex.z-dn.net/?f=F%3D15%5Ctimes%2010%5E%7B-6%7D%5B%284.53i%29%5Ctimes%20%280.08%29%5C%20j%5D)


So, the force acting on the charge is
and is moving in positive z axis. Hence, this is the required solution.
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Answer:15metre per second squ
I
Explanation:
acceleration=(final velocity-initial velocity)÷t
acceleration=(180-120)÷4
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Answer:
A
Explanation:
i feel it is correct
in the sense that it dwpends on the presence of 4different antigens
Force of Gravity depends on:
1) Masses of the object
2) Distance between the object
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Two or three i believe to be the answer