all T-Mobile 10:32 PM flipitphysics.com FlipltPhyscs Copy of Electricity &Magnet

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all T-Mobile 10:32 PM flipitphysics.com FlipltPhyscs Copy of Electricity &Magnetism; I University Physics (calculus-bas Homework: Magnetism inMagnetie Feld 1 A charged particle of mass m . 6.5X10-8 kg, moving with constant velocity in the y-direction enters a +x"@ region containing a constant magnetic field B .5T aligned with the positive z-axis as shown. The d particle enters the region at (x,y)-(0.97 m, 0) and leaves the region at (x,y-0. t . 830 is after it entered theregion.0.97 matinet-830 -ile With what speed v did the particle enter the region containing the magnetic field? what is Fx, the x-component of the force on the particle at a time t 1 . 276.7 us after it entered the region containing the magnetic field What is Fy, the y component of the force on the particle at a time t 276.7 us after it entered the region containing the magnetic field What is q, the charge of the particle? Be sure to include the correct sign If the velocity of the incident charged particle were doubled, how would B have to change (keeping all other parameters constant) to keep the trajectory of the particle the same? Oincrease B by a factor of 2 O Increase B by less than a factor of 2 O Decrease B by less than a factor of 2 Decrease B by a factor of 2 There is no change that can be made to B to keep the trajectory the "Below is some space to write notes on this problem.

Explanation / Answer

1) speed v the particle enter the region containing the magnetic field, v= s/t = (d*pi/2)/t =( 0.97*3.14/2)*830e-6= 1834.8192 m/s

2)si=d*theta1=v*t1---> theta1= v*t1/d

F=m*v^2/d-----> Fx= -m*(v^2/d) *cos theta1

Theta1= 1834.8192*276.7e-6/0.97 =0.523

Fx= -6.5e-8*(1834.8192)^2 * cos (0.523)/0.97= -0.1954 N

3) Fy= -m*(v^2/d) sin theta1 = 6.5e-8*(1834.8192)^2 * sin (0.523)/0.97= -0.1127 N

4) q=- m*v/d*B= (6.5e-8*1834.8192)/(0.97*1.5)= -81.96 C

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