Model a magnetic field exerts a magnetic force on a

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Model: A magnetic field exerts a magnetic force on a length of current-carrying wire. We ignore gravitational effects, and focus on the B effects. Visualize: Please refer to Figure P33.67. The figure shows a wire in a magnetic field that is directed out of the page. The magnetic force on the wire is therefore to the right and will stretch the springs. Solve: In static equilibrium, the sum of the forces on the wire is zero: F B + F sp 1 + F sp 2 = 0 N ILB + ( k Δ x ) + ( k Δ x ) ( )( ) ( )( ) 2 10 N/m 0.01 m 2 2.0 A 0.20 m 0.5 T k x I LB Δ = = =
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33.68. Model: The bar is a current-carrying wire in a perpendicular uniform magnetic field. The current is constant. Visualize: Please refer to figure P33.68. Solve: (a) The right-hand rule as described in section 33.8 requires the current to be into the page. (b) The net force on the bar is , F IlB = and is constant throughout the motion. The acceleration of the bar is thus . F IlB m m = Using constant acceleration kinematics, ( ) 2 2 2 0 2 2 0 m/s 2 f f IlB IlBd v v a s d v m m = + Δ = + =
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33.69. Model: A magnetic field exerts a magnetic force on a length of current-carrying wire. Visualize: Please refer to Figure P33.69. Solve: The above figure shows a side view of the wire, with the current moving into the page. From the right- hand rule, the magnetic field B G points down to give a leftward force on the current. The wire is hanging in static equilibrium, so net mag 0 G F F F T = + + = G G G G N. Consider a segment of wire of length L . The wire’s linear mass density is μ = 0.050 kg/m, so the mass of this segment is m = μ L and its weight is F G = mg = μ Lg. The magnetic force on this length of wire is F mag = ILB . In component form, Newton’s first law is ( ) net mag sin sin 0 N x F T F T ILB θ θ = = = sin T ILB θ = ( ) net cos cos 0 N G y F T F T Lg θ θ μ = = = cos T Lg θ μ = Dividing the first equation by the second, sin tan cos T ILB IB T Lg g θ θ θ μ μ = = = ( ) ( ) 2 0.050 kg/m 9.8 m/s tan10 tan 0.0086 T 10 A g B I μ θ ° = = = The magnetic field is ( ) 8.6 mT, down B = G .
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33.70. Model: The wire will float in the magnetic field if the magnetic force on the wire points upward and has a magnitude mg , allowing it to balance the downward gravitational force. Visualize: Solve: Each lower wire exerts a repulsive force on the upper wire because the currents are in opposite directions. The currents are of equal magnitude and the distances are equal, so 1 2 . F F = Consider segments of the wires of length L . Then the forces are 2 0 1 2 2 LI F F d μ π = = The horizontal components of these two forces cancel, so the net magnetic force is upward and of magnitude 2 0 mag 1 cos30 2 cos30 LI F F d μ π ° = ° = In equilibrium, this force must exactly balance the downward weight of the wire. The wire’s linear mass density is μ = 0.050 kg/m, so the mass of this segment is m = μ L and its weight is w = mg = μ Lg . Equating these gives 2 0 cos30 LI Lg d μ μ π ° = ( ) ( ) ( ) ( ) 2 7 0 0.050 kg/m 9.8 m/s 0.040 m 238 A cos30 4 10 T m/A cos30 g d I π μ π μ π = = = ° × °
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33.71.
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  • Winter '10
  • E.Salik
  • Current, Magnetic Field, ΔS

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