Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 31, Problem 31.83CP

Review. The bar of mass m in Figure P30.51 is pulled horizontally across parallel, frictionless rails by a massless string that passes over a light, frictionless pulley and is attached to a suspended object of mass M. The uniform upward magnetic field has a magnitude B, and the distance between the rails is . The only significant electrical resistance is the load resistor R shown connecting the rails at one end. Assuming the suspended object is released with the bar at rest at t = 0, derive an expression that gives the bar’s horizontal speed as a function of time.

Figure P30.51

Chapter 31, Problem 31.83CP, Review. The bar of mass m in Figure P30.51 is pulled horizontally across parallel, frictionless

Expert Solution & Answer
Check Mark
To determine
The expression for horizontal speed of the bar as a function of time.

Answer to Problem 31.83CP

The expression for horizontal speed of the bar as a function of time is MgRB2l2(1eB2l2t(m+M)R) .

Explanation of Solution

Given info: Mass of bar is m , mass of object is M , upward magnetic field is B , distance between the rails is l and resistance of load resistor is R .

The emf induced in the bar can be given as,

ε=Blv

Here,

ε is the emf induced in the bar.

B is the uniform magnetic field

l is the distance between the rails.

v is the horizontal speed of bar.

The current induced in the bar can be given as,

I=εR

Here,

I is the current induced in the bar.

R is the resistance of the load resistor.

Substitute Blv for I in the above equation,

I=BlvR

The force induced in the bar due to magnetic field can be given as,

F=IBl

Here,

F is the force induced in the bar.

Substitute BlvR for I in the above equation,

F=(BlvR)Bl=B2l2vR (1)

The force due to weight can be given as,

W=Mg (2)

Here,

M is the mass of object.

g is the acceleration due to gravity.

W is the force due to weight.

As, force due to magnetic field and force due to weight will act in opposite direction, the net force acting on the bar can be given by subtracting equation (2) from equation (1),

Fnet=MgB2l2vR (3)

Here,

Fnet is the net force acting on the bar.

The net force can also be given as,

Fnet=(m+M)dvdt

Substitute (m+M)dvdt for Fnet in the equation (3),

(m+M)dvdt=MgB2l2vRdvdt+B2l2v(m+M)R=M(m+M)g (4)

The equation (4) is a linear differential equation of the form,

dvdt+Pv=Q

Here,

P is B2l2(m+M)R and a function of t .

Q is M(m+M)g and a function of t .

The integrating factor for the equation (4) can be given as,

I.F.=ePdt=eB2l2(m+M)Rdt=eB2l2t(m+M)R

Here,

I.F. is the integrating factor.

The solution for the differential equation is,

v(I.F.)=Q(I.F.)dt+C

Here,

C is the constant.

Substitute eB2l2t(m+M)R for I.F. in the above equation,

veB2l2t(m+M)R=M(m+M)geB2l2t(m+M)Rdt+CveB2l2t(m+M)R=(M(m+M))((m+M)RB2l2)geB2l2t(m+M)R+CveB2l2t(m+M)R=MgRB2l2eB2l2t(m+M)R+C (5)

Apply boundary condition, v=0 at t=0 in the above equation,

0=MgRB2l2+CC=MgRB2l2

Substitute (MgRB2l2) for C in the equation (5),

veB2l2t(m+M)R=MgRB2l2eB2l2t(m+M)RMgRB2l2veB2l2t(m+M)R=MgRB2l2(eB2l2t(m+M)R1)v=MgRB2l2(1eB2l2t(m+M)R)

Thus, the expression for speed of the bar is MgRB2l2(1eB2l2t(m+M)R) .

Conclusion:

Therefore, the expression for horizontal speed of the bar as a function of time will be MgRB2l2(1eB2l2t(m+M)R) .

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Chapter 31 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

Ch. 31 - The bar in Figure OQ31.6 moves on rails to the...Ch. 31 - A bar magnet is held in a vertical orientation...Ch. 31 - What happens to the amplitude of the induced emf...Ch. 31 - Two coils are placed near each other as shown in...Ch. 31 - A circuit consists of a conducting movable bar and...Ch. 31 - Two rectangular loops of wire lie in the same...Ch. 31 - In Section 7.7, we defined conservative and...Ch. 31 - A spacecraft orbiting the Earth has a coil of wire...Ch. 31 - In a hydroelectric dam, how is energy produced...Ch. 31 - A bar magnet is dropped toward a conducting ring...Ch. 31 - A circular loop of wire is located in a uniform...Ch. 31 - A piece of aluminum is dropped vertically downward...Ch. 31 - Prob. 31.7CQCh. 31 - When the switch in Figure CQ31.8a is closed, a...Ch. 31 - Prob. 31.9CQCh. 31 - A loop of wire is moving near a long, straight...Ch. 31 - A flat loop of wire consisting of a single turn of...Ch. 31 - An instrument based on induced emf has been used...Ch. 31 - Transcranial magnetic stimulation (TMS) is a...Ch. 31 - A 25-turn circular coil of wire has diameter 1.00...Ch. 31 - A circular loop of wire of radius 12.0 cm is...Ch. 31 - A circular loop of wire of radius 12.0 cm is...Ch. 31 - Prob. 31.7PCh. 31 - A strong electromagnet produces a uniform magnetic...Ch. 31 - A 30-turn circular coil of radius 4.00 cm and...Ch. 31 - Scientific work is currently under way to...Ch. 31 - An aluminum ring of radius r1 = 5.00 cm and...Ch. 31 - An aluminum ring of radius r1 and resistance R is...Ch. 31 - Prob. 31.13PCh. 31 - A coil of 15 turns and radius 10.0 cm surrounds a...Ch. 31 - A square, single-turn wire loop = 1.00 cm on a...Ch. 31 - A long solenoid has n = 400 turns per meter and...Ch. 31 - A coil formed by wrapping 50 turns of wire in the...Ch. 31 - When a wire carries an AC current with a known...Ch. 31 - A toroid having a rectangular cross section (a =...Ch. 31 - Prob. 31.20PCh. 31 - A helicopter (Fig. P30.11) has blades of length...Ch. 31 - Use Lenzs law 10 answer the following questions...Ch. 31 - A truck is carrying a steel beam of length 15.0 in...Ch. 31 - A small airplane with a wingspan of 14.0 m is...Ch. 31 - A 2.00-m length of wire is held in an eastwest...Ch. 31 - Prob. 31.26PCh. 31 - Figure P31.26 shows a lop view of a bar that can...Ch. 31 - A metal rod of mass m slides without friction...Ch. 31 - A conducting rod of length moves on two...Ch. 31 - Prob. 31.30PCh. 31 - Review. Figure P31.31 shows a bar of mass m =...Ch. 31 - Review. Figure P31.31 shows a bar of mass m that...Ch. 31 - The homopolar generator, also called the Faraday...Ch. 31 - Prob. 31.34PCh. 31 - Review. Alter removing one string while...Ch. 31 - A rectangular coil with resistance R has N turns,...Ch. 31 - Prob. 31.37PCh. 31 - An astronaut is connected to her spacecraft by a...Ch. 31 - Within the green dashed circle show in Figure...Ch. 31 - Prob. 31.40PCh. 31 - Prob. 31.41PCh. 31 - 100-turn square coil of side 20.0 cm rotates about...Ch. 31 - Prob. 31.43PCh. 31 - Figure P30.24 (page 820) is a graph of the induced...Ch. 31 - In a 250-turn automobile alternator, the magnetic...Ch. 31 - In Figure P30.26, a semicircular conductor of...Ch. 31 - A long solenoid, with its axis along the x axis,...Ch. 31 - A motor in normal operation carries a direct...Ch. 31 - The rotating loop in an AC generator is a square...Ch. 31 - Prob. 31.50PCh. 31 - Prob. 31.51APCh. 31 - Suppose you wrap wire onto the core from a roll of...Ch. 31 - A circular coil enclosing an area of 100 cm2 is...Ch. 31 - A circular loop of wire of resistance R = 0.500 ...Ch. 31 - A rectangular loop of area A = 0.160 m2 is placed...Ch. 31 - A rectangular loop of area A is placed in a region...Ch. 31 - Strong magnetic fields are used in such medical...Ch. 31 - Consider the apparatus shown in Figure P30.32: a...Ch. 31 - A guitars steel string vibrates (see Fig. 30.5)....Ch. 31 - Why is the following situation impossible? 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In Figure P30.42, a uniform magnetic field...Ch. 31 - An N-turn square coil with side and resistance R...Ch. 31 - A conducting rod of length moves with velocity v...Ch. 31 - The magnetic flux through a metal ring varies with...Ch. 31 - A rectangular loop of dimensions and w moves with...Ch. 31 - A long, straight wire carries a current given by I...Ch. 31 - A thin wire = 30.0 cm long is held parallel to...Ch. 31 - Prob. 31.79CPCh. 31 - An induction furnace uses electromagnetic...Ch. 31 - Prob. 31.81CPCh. 31 - A betatron is a device that accelerates electrons...Ch. 31 - Review. The bar of mass m in Figure P30.51 is...
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