a. What is the net magnetic field at point C? Give both its magnitude and its di

Question

a. What is the net magnetic field at point C? Give both its magnitude and its direction with respect to the axes shown.

b. Supose that a particle with a mass of 2.07E-8 kg and a net chrage of -1.5E-6 C is passing through point C with a velocity of magnitude 3.4E3 m/s pointed at angle of 15 degrees above the +x axis.

i. What is the magnetic force felt by the particle? Give both the magnitude and direction of the force.

ii. What is the instantaneous acceleration of the particle? Again, give both the magnitude and the direction of the force.

Explanation / Answer

magnetic field Due to A is Bba along y axis

Bay = uoI/2pia

Bay = (4pi *10^-7 * 4.5)/(2pi* a)

here

Cos theta = AB/CB

CB = AB/cos 30

CB = 0.04/cos 30

CB = 0.046 m

sin theta = AC/CB

AC = CB * sin theta

AC = 0.045 * sin 30

AC = 0.0225 m

Bay = (4pi *10^-7 * 4.5)/(2pi* 0.0225)

Bay = 4*10^-5 Tesla

Bax = 0

magnetic field due to B,

Bbx = uo*I*cos30/(2pi*a))

so

Bbx = 4pi *10^-7 *cos 30 * 0.45/(2pi* 0.046)

Bbx = 1.7 *10^-6 Tesla

Bby = uoIsin theta/2pia

Bby = (4pi *10^-7 * 4.5 * sin 30)/(2pi *0.0225)

Bby = 2*10^-5 Tesla

Bx = 0 + 1.7 *10^-6 = 1.7 *10^-6 T

By = 0.4 *10^-6 + 0.2*10^-6 = 0.6 *10^-6 T

Bnet^2 = Bx^2 + By^2

Bnet^2 = 1.7^2 + 0.6^2

Bnet at C = 1.802 *10^-6 Tesla

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b. magnetic force F = q vB sin theta

Fnet = 1.5*10^-6 * 3.4 *10^3 * 1.8 *10^-6 * sin 15

Fnet = 2.375 *10^-9 N

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i)

direction of force = tan theta = BY/Bx

tan theta = 0.6/1.7

theta = 19.44 deh

angle of Deflection is 19.44 +15 = 34.33 deg

---------------
ii)

acccleration a = F/m

a = 2.375 *10^-9/(2.07 *10^-8)

a = 0.114 m/s^2

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