KLscilevothma
May25-03, 05:02 AM
It's Q1 from the physics contest (high school) that I attended today. I would like to check my answers here though I don't think I got the correct answers. Anyway, here it goes.
As shown in the figure, a conducting slope frame with inclining angle %theta; is placed in a uniform magnetic field B pointing upwards. A conducting rod of length L, mass M, resistance R is moving down the frame at steady speed v. (Hint the emf of a rod moving in magnetic field is BLvp, where vp is the velocity component perpendicular to the field. If the rod carries electric current "I" which is flowing in a direction perpendicular to the field, then the magnetic force on the rod is BLI.)
a) Find v in terms of θ, L, M, R, B, and any other physics constants you think appropriate.
b) If the direction of B is reversed, what will be the direction of the steady velocity when the rod is released from rest?
c) If R=[oo] , describe the motion of the rod after it is released from rest.
Here's my approach:
a) emf = BLvcosθ = IR ..........(1)
since the rod is moving down at steady speed,
therefore Magnetic force(which acts up) = weight component, mgsinθ, which acts down. (I skip the right hand rule, left hand rule part here)
ie BLIcosθ= Mg sinθ (where g is the acceleration due to gravity)
I = Mgsinθ/(BLcosθ) .......... (2)
sub (2) in (1)
we got
BLvcosθ = IR = Mgsinθ/(BLcosθ)R
therefore v = MgRsinθ/[(BL)2cos2θ]
b) If the direction of B is reversed, direction of current flows in the rod reverse, the magnetic force acting on the rod still acts up, therefore the direction of the steady velocity doesn't change. ie The rod goes down.
c) If R = [oo], current flows through the rod tends to zero. Therefore there is hardly any magnetic force on the rod which acts up. So the rod goes down, where the acceleration is gsinθ
As shown in the figure, a conducting slope frame with inclining angle %theta; is placed in a uniform magnetic field B pointing upwards. A conducting rod of length L, mass M, resistance R is moving down the frame at steady speed v. (Hint the emf of a rod moving in magnetic field is BLvp, where vp is the velocity component perpendicular to the field. If the rod carries electric current "I" which is flowing in a direction perpendicular to the field, then the magnetic force on the rod is BLI.)
a) Find v in terms of θ, L, M, R, B, and any other physics constants you think appropriate.
b) If the direction of B is reversed, what will be the direction of the steady velocity when the rod is released from rest?
c) If R=[oo] , describe the motion of the rod after it is released from rest.
Here's my approach:
a) emf = BLvcosθ = IR ..........(1)
since the rod is moving down at steady speed,
therefore Magnetic force(which acts up) = weight component, mgsinθ, which acts down. (I skip the right hand rule, left hand rule part here)
ie BLIcosθ= Mg sinθ (where g is the acceleration due to gravity)
I = Mgsinθ/(BLcosθ) .......... (2)
sub (2) in (1)
we got
BLvcosθ = IR = Mgsinθ/(BLcosθ)R
therefore v = MgRsinθ/[(BL)2cos2θ]
b) If the direction of B is reversed, direction of current flows in the rod reverse, the magnetic force acting on the rod still acts up, therefore the direction of the steady velocity doesn't change. ie The rod goes down.
c) If R = [oo], current flows through the rod tends to zero. Therefore there is hardly any magnetic force on the rod which acts up. So the rod goes down, where the acceleration is gsinθ