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A particle with charge 7.80 muC is movin...

A particle with charge `7.80 muC` is moving with velocity `v = - (3.80 xx 10^3 m/s)hatj`. The magnetic force on the particle is measured to be `F = + (7.60 xx 10^-3 N)hati - (5.20 xx 10^-3 N)hatk`
(a) Calculate the components of the magnetic field you can find from this information.
(b) Are the components of the magnetic field that are not determined by the measurement of the force? Explain.
(c) Calculate the scalar product `B .F`. What is the angle between `B` and `F`

Text Solution

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a. `F=q(vxxB)`
or `[(7.6xx10^-3)hati-(5.2xx10^-3)hatk]`
`=(7.8xx10^-6)[(-3.8xx10^3)hatj(B_xhati+B_yhatj+B_2hatk)]`
`:.(7.8xx10^-6)(3.8xx10^-3)(B_x)`
`=(-5.2xx10^-3)`
`:.B_x=-0.175T`
Similarly
`(7.8xx10^-6)(-3.8xx10^3)(B_z)`
`=7.6x10^-3)`
or `B_z=-0.256T`
c. From the property of cross product.
`F` is always perpendicular to `B`.
Hence, `F.B=0`
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