Answer for a) down B) 1.739x10-4 C)6.55 micrometers D)1800NM What are the in-dep

Question

Answer for a) down B) 1.739x10-4 C)6.55 micrometers D)1800NM What are the in-depth steps in solving this? Quesm electron and a proton traveling at the same velocity (100m/s to the right) enter a uniform magnetic field as shown. The two particles experience a magnctic force while in the magnetic field and exit the field traveling to the lef·The electron exits the field a height h away from the point where it entered while the proton exits a distance (h+ 12mm) from the point where it entered Ouestion s e, p (a) Will the electron curve upwards or downwards? (b) Calculate the distance between the entrance point and exit pointo (c) Calculate the magnetic field (d) A rectangular closed loop is suspended from a spring as shown. A current of 2A runs through the closed loop. The bottom half of the loop hangs in a uniform magnetic field of 3T, while the top of the loop experiences no magnetic field. At some time, the current is slowly turned off and the loop moves 1 mm to its new equilibrium position. What is the spring constant of the spring? The loop is 15 cm tall and 30 cm wide.

Explanation / Answer

part a:

magnetic force on a moving charge is given by

q*cross product of v and B

where q=charge

v=velocity

B=magnetic field

for an electron , q is negative.

v is along +ve x axis, and B is along -ve z axis.

then cross product of v and B is along +ve y axis.

with q being negative, force is along -ve y axis.

hence force is in downward direction

part b:

distance between entry point and exit point is diameter of the path traced by electron

let radius of the path be r meters.

then balancing magnetic force with centripetal force:

q*v*B=m*v^2/r

==>r=m*v/(q*B)

as given distance between electron and proton exit point is 12 mm

hence difference between their diameter is 12 mm

==>difference between their radii is 6 mm

==>(v/(q*B))*(mass of proton-mass of electron)=6 mm

==>(100/(1.6*10^(-19)*B)*(1.67*10^(-27)-9.1*10^(-31))=0.006

==>B=1.7386*10^(-4) T

then radius for electron=r=m*v/(q*B)

=(9.1*10^(-31)*100)/(1.6*10^(-19)*1.7386*10^(-4))=3.2713*10^(-6) m

then distance between entry and exit point=diameter of the path

=2*3.2713*10^(-6) m

distance between entry and exit=6.5426 micrometer

part c:

as found in part b,

magnetic field=1.7386*10^(-4) T

part d:

let spring constant be k.

magnetic force+weight of the loop is balanced by the spring force.

let original spring elongation be x0 m

then k*x0=magnetic force+weight…(1)

after current is turned off, magnetic force is 0.

so elongation will be x0-1 mm=x0-0.001 m

then k*(x0-0.001)=weight…(2)

subtracting equation 2 from equation 1,

k*0.001=magnetic force..(3)

magnetic force=current*cross product of length and magnetic field vector

force on the right vertical side will be balanced by force on the left vertical side.

so net force=force on the bottom horizontal portion

=current*length*magnetic field

==>k*0.001=2 A*0.3 m*3 T

==>k=2*0.3*3/0.001

=1800 N.m

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