At a particular instant, a muon is in a magnetic field, with its velocity perpen

Question

At a particular instant, a muon is in a magnetic field, with its velocity perpendicular to that field. At that instant, it experiences an acceleration of 5.5e+14 m/s2 straight down. This is possible if its speed is 1 m/s and the magnetic field is _____ T.

Select possible pair(s) of directions:

Velocity in -z and magnetic field in +x

Velocity in +x and magnetic field in -z

Velocity in -z and magnetic field in -x

Velocity in +x and magnetic field in +z

Velocity in +y and magnetic field in +z

Velocity in +z and magnetic field in +y

Explanation / Answer

charge of muon=-1.6*10^(-19) C

mass of muon=1.8843*10^(-28) kg

force due to magnetic field=charge*speed*magnetic field*sin(theta)

where theta is angle between velocity and magnetic field

here theta=90 degrees

force=mass*acceleration

hence mass*acceleration=charge*speed*magnetic field

==>1.8843*10^(-28)*5.5*10^14=1.6*10^(-19)*1*magnetic field

==>magnetic field=(1.8843*10^(-28)*5.5*10^14)/(1.6*10^(-19))=6.4773*10^5 T

combinations are:

direction of force is -ve y axis.

as charge is -ve, cros product of velocity and magnetic field is along +ve y axis

so possible combinations are:

velocity in -z and magnetic field in -x

velocity in +x and magnetic field in -z

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