A proton travels with a speed of 3.00 times 10 6 m/s at an angle of 44.0 degree

Question

A proton travels with a speed of 3.00 times 10 6 m/s at an angle of 44.0 degree with the direction of a magnetic field of 0.400 T in the +y direction. What is the magnitude of the magnetic force on the proton? What is the acceleration? m/s2 A current of 12.0 mA is maintained in a Single circular loop of 2.00 m circumference. A magnetic field of 0.790 T is directed parallel to the plane of the loop. Calculate the magnetic moment of the loop. mA-m2 What is the magnitude of the torque exerted by the magnetic field on the loop? mN-m A velocity selector of electric and magnetic fields described by the expressions E rightarrow = E K^ and B rightarrow = B j^, with B = 0.0100 T. Find the value of E such that a 730 -eV electron moving along the negative X axis is undeflected. kV/m

Explanation / Answer

15)

i=12 mA and 2*pi*r=2m

==> r=0.318 m

a)

M=i*A

=(12*10^-3)*(pi*r^2)

=(12*10^-3)*(3.14*0.318^2)

=3.822*10^-3 A.m^2

=3.822 mA.m^2

b)

Torque

T=MB*cos(theta)

=3.822*0.79*cos(0)

=3.019 mN.m

16)

Fe=FB

q*E=q*v*B

E=v*B

and

1/2*m*v^2=U

v=sqrt(2*U/m)

v=sqrt(2*730*1.6*10^-19/9.1*10^-31)

v=16.022*10^6 m/sec

now

E=v*B

E=(16.022*10^6)*0.01

=160.22*10^3 V/m .

=160.22 kV/m

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