A trip through magnetic fields

A positively charged particle, which was initially moving in the + y +y direction on the x y xy -plane, passes through points P , Q and R , as shown above. The magnetic field intensities of area I and area II are 2 B 2B and B , B, respectively, and P and Q have the same distance from the origin O . If it takes T 0 T_0 for the charged particle to move from P to Q , how long will it take for it to move from Q to R ?

4 3 T 0 \frac{4}{3}T_0 2 3 T 0 \frac{2}{3}T_0 3 4 T 0 \frac{3}{4}T_0 T 0 T_0

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2 solutions

Himanshu Arora
Jun 20, 2014

Radius, R of the trajectory (circle) of a charged particle, (m,q) moving with speed v in magnetic field, B is given by R = m v q B R=\frac{mv}{qB} . From this we can see that OQ is half the radius of the original arc QR. Thus the particle traverses an angle of π / 3 \pi/3 about the center of revolution. Also the time period of circular motion is given by T = 2 π m q B T= \frac{2\pi m}{qB} . Combine the two to get 4 T 3 \boxed{\frac{4T}{3}}

Kanishkan R
Dec 7, 2014

in the field B particle has to travel more distance than that in 2B,since velocity does not affect due to magnetic field the time the particle to cover the larger distance will be greater than To,neglecting all the smaller value we can find the answer,its not the appropriate method but still can save some time......:)

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