Question

1.-

A)

A proton of cosmic ray in interstellar space has an energy of 7.50 MeV and executes a circular orbit of radius equal to that of Mercury's orbit around the Sun (5.80 x 1010 m). What is the magnetic field in that region of space?

a

9.648e-031 T

b

2.295e + 010 T

c

None of the above

d

2.699e-012 T

e

6.822e-012 T

B)

A circular coil of 55.00 turns and 1.00 cm radius can be oriented in any direction in a uniform magnetic field that has a magnitude of 1.00 T. If the coil carries a current of 35.00 mA, find the magnitude of the maximum possible torque exerted on the coil

a

2.42e-003 N.m

b

None of the above

c

6.048e-004 N.m

d

6.048e-004 N.m

e

1.999e-007 N.m

Answer 1

Problem1A

Given : Interstellar space energy = 7.50 MeV

Solution:

Since it is circular path, the particle moves perpendicular to the magnetic field(B), hence the magnetic force supplies the centripetal acceleration ,

therfore F_C=F_B

  mv²/r = qvB

B = mv/qr

the linear momentum of the proton is given by p = mv =sqrt (2mKE)

KE (kinetic energy) of the proton = 7.50MeV

KE = (7.50*10^6eV) (1.60*10^-19J / 1eV) = 1.2*10^-12 J

Determining the magnetic field B

B = mv/qr = p/qr = sqrt(2mKE)/qr

where mass of the proton = 1.67*10^-27kg

charge of a proton = 1.60*10^-19C

   B = sqrt (2*(1.67*10^-27Kg)*(1.2*10^-12J)) / (1.60*10^-19C)(5.80*10^10m)

B = 6.822*10^12 T

OPTION e IS CORRECT

Problem 1B

Given :

No. of turns = 55

radius= 1cm= 1*10^-2m

magnitude = 1T

current, I= 35mA = 35*10^-3A

Solution :

The torque on a cureent loop in a magnetic field is = BIAN sin

The maximum torque occurs when the field is directly parallel to the plane of the loop

( = 90 degree)

  _max = (1T)(35*10^-3 A)(*(1*10^-2 m)^2) (55) sin90

_max = 6.048*10^-4Nm