Practice test problems for physics 202 - Need help! very charged tr particle of

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Practice test problems for physics 202 - Need help!

very charged tr particle of charge IC and mass m 2kg enters a region nitorm magnetic field B = 2T. The magnetic field is per he particl pendicular to the velocity of Assume that this isa situation where the particle will experience e u = 4 niform circular motion governed by quB- Find the radius R. 25 points) Velocity Selector n a velocity selector such as this one, there is a magnetic field and an electric field etween two plates of a parallel plate capacitor. Suppose positively charged particles re moving from left towards the right through the velocity selector. Only particles that ave a specific velocity will go through undeflected. These particles experience no net orce when they are between the plates, ie, FE = FB. That is to say, the electric force alances the magnetic force. Recall that FE 9E and in this case F:= quB. (a) (10 points) If the velocity and magnetic fields are as shown on the figure on the board. Draw a diagram showing the way in which a battery must be connected in order to set up an electric force that opposes the magnetic force. (Hint: use the right hand rule to determine the direction of the magnetic force then draw arrows going in the opposite direction for the electric force. Since the direction of the electric force is the same as the direction of the electric field use this to figure out how the battery must be connected to give this kind of electric field) (b) (15 points) Suppose the particle is moving with velocity u = 8, the magnetic field is B = 3T the separation d between the plates is d = , what must the voltage V of the battery be in order for this particle to pass through the plates undeflected? (Hint: -)

Explanation / Answer

a) v = left to right (along i vector )

q = +v3

B = inside the page ( along -z vector)    ( and j will be along upward)

F = q(v X B) = + ( i X -z ) = -j

so magnetic force will be along downward.

so electric field have t o be upward.

up plate will be -ve and lower plate will be +ve.

so option (ii)

b) F_magnetic = qvB = q(8 x 3) = 24q

F_ele = qE = qV/d = q2V

equating them ,

24q = 2Vq

V = 24 Volt

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