An electron has a velocity of 1.00 km/s (in the positive x direction) and an acc

Question

An electron has a velocity of 1.00 km/s (in the positive x direction) and an acceleration of 2.00 times 10 12 m/s2 ( in the positive z direction) in uniform electric and magnetic fields. If the electric field has a magnitude of strength of 25.0 N/C (in the positive z direction), determine the following components of the magnetic field. If a component cannot be determined, enter 'undetermined'. BX = T BY = T BZ = T A proton moves with a velocity of v rightarrow = (5i^ - 4j^ + k^) m/s in a region in which the magnetic field is B rightarrow = (i^ + 2j^ - 3k^) T. What is the magnitude of the magnetic force this charge experiences? N A wire 1.70 m long carries a current of 33.8 mA when it is connected to a battery. The whole wire can be arranged as a single loop with the shape of a circle, a square, or an equilateral triangle. The whole wire can be made into a flat, compact, circular coil with N turns. Calculate the magnetic moments in each case.

Explanation / Answer

when a harged particle is in both electrical and magnetic fields it experiences both magnetic force and eletric force

electric force Fe = q*E k

magnetic force Fb = q*(vcrossB) = q * (Vxi x (Bxi + Byj + Bzk )

Fb = q*Vx*By k - q*Vx*Bz j

net force Fnet = Fe + Fb

Fnet = q*E k + q*Vx*By k - q*Vx*Bz j

Fnet = - q*Vx*Bz j + q*(E + vx*by) k

but net force on a p article is Fnet = m*az

m*az = - q*Vx*Bz j + q*(E + vx*by) k

az = - q*Vx*Bz/m j + q*(E + vx*By)/m k

comparing j

az = 0========> Bz = 0

comparing k

q*(E + vx*By)/m = az

By = (m*az - Eq)/(q*vx)

By = ((9.1e-31*2e12)-(25*1.6e-19))/(1.6e-19*1000) = -0.013625 T

Bx undetermined

Bx = undetermined

By = 0.013625 T

Bz = 0

+++++++++++++++++++

Fb = q*(v cross B)

Fb = 1.6e-19*(10i + 16j + 14 k)

Fb = 16*10^-19 i + 9.6*10^-19 N + 22.4*10^-19 N

++++++

for circle

2*pi*r = L

r = L/(2*pi) = 1.7/(6.28) = 0.27 m

A1 = pi*r^2 = 0.23 m^2

u = I*A1 = 33.8e-3*0.23 = 0.007774 Am^2

for aquare

4r = L

r = L/4

A2 = r^2 = (L/4)^2 = 0.180625 m^2

u = I*A2 = 6.105125*10^-3 Am^2

+++++++++++

for equilateral traingle

3r = L

r = L/3

A3 = sqrt(3)*r^2/4 = 0.139

u = I*A3 = 4.6982*10^-3 Am^2

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