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A point chanrge of magnitude q=4.5nC is ...

A point chanrge of magnitude `q=4.5nC` is moving with speed `v=3.6xx10^7ms^-1` parallel to the x-axis along the line `y=3m`. Find the magnetic field at the origin produced by this charge when the charge is at the point `x=-4 m`, `y=3m`, as shown in Fig.

Text Solution

Verified by Experts

The magnetic field is given by
`vecB=(mu_0)/(4pi) (qvecvxxhatr)/(r^2)`, with `vecv=vhati`
`vecr=(4hati-3hatj)m implies vecr=sqrt(4^2+3^2)m=5m`
Unit vector in the direction of `hatr=(vecr)/(vecr)=(4hati-3hatj)/5=0.8hati-0.6hatj`
`vecvxxhatr=(vhati)xx(0.8hati-0.6hatj)=-0.6vhatk`
`vecB=(mu_0)/(4pi) (qvecvxxhatr)/(r^2)=(mu_0)/(4pi) (q(-0.6vhatk))/(r^2)`
`=-(10^-7)((4.5xx10^-19)(0.6)(3.6xx10^7))/(5^2)hatk`
`=-3.89xx10^-10 Thatk`
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