# Work and Heat Transfer in a Magnetic Field: Conduction Rod on Conducting Rails

• Titan97
In summary: The net work done is zero because the work done by magnetic field is equal to the work done by the feedback force. In summary, a conduction rod with initial velocity u is placed in a circuit with a resistor and a constant uniform magnetic field. As the rod moves, an emf is induced and a retarding force is produced due to the magnetic field. The work done by the magnetic field is equal to the work done by the feedback force from the circuit, resulting in a net work done of zero. The heat generated in the circuit is a result of the conversion of kinetic energy to electrical energy and back to mechanical energy. The rate of energy consumption is determined by the resistance of the circuit.
Titan97
Gold Member

## Homework Statement

A conduction rod (of mass ##m## and length ##l##) was placed on two smooth conducting rails rails connected by a resistor as shown: (the circuit is placed in ##XY##-plane

A constant uniform magnetic field is switched on along ##-Z## direction with magnitude ##B##
The rod is given an initial velocity ##u##. Find the work done by magnetic field when the velocity of rod becomes ##v## and the heat energy liberated till that time.

## Homework Equations

##V_{induced}=Blv##
##F=ilB##

## The Attempt at a Solution

An emf is induced in the loop since flux due to ##B## increases. A retarding force ##F=ilB## acts.
$$i=\frac{Blv}{R}$$
$$F=\frac{B^2l^2v}{R}=-m\frac{dv}{dt}$$ (since velocity decreases as time increases)

Since the force due to magnetic field is the only force acting on the rod, its kinetic energy change is because of this force. But if there is current through a resistor, then there will be heat loss. But if all the work done is used for changing kinetic energy, where does heat come from?

Titan97 said:
But if there is current through a resistor, then there will be heat loss.

Not sure, but I think you answered your own question. If you know the current through the resistor, and its resistance, can you calculate how much heat is dissipated?

Titan97 said:

## Homework Statement

A conduction rod (of mass ##m## and length ##l##) was placed on two smooth conducting rails rails connected by a resistor as shown: (the circuit is placed in ##XY##-plane
View attachment 93629
A constant uniform magnetic field is switched on along ##-Z## direction with magnitude ##B##
The rod is given an initial velocity ##u##. Find the work done by magnetic field when the velocity of rod becomes ##v## and the heat energy liberated till that time.

## Homework Equations

##V_{induced}=Blv##
##F=ilB##

## The Attempt at a Solution

An emf is induced in the loop since flux due to ##B## increases. A retarding force ##F=ilB## acts.
$$i=\frac{Blv}{R}$$
$$F=\frac{B^2l^2v}{R}=-m\frac{dv}{dt}$$ (since velocity decreases as time increases)

Since the force due to magnetic field is the only force acting on the rod, its kinetic energy change is because of this force. But if there is current through a resistor, then there will be heat loss. But if all the work done is used for changing kinetic energy, where does heat come from?
Interesting question! Here's what I think about the situation..
Kinetic energy given to the rod will be converted into electrical energy. Motional emf will be induced in the rod. This will drive current through the circuit. Heat will be a part of the kinetic energy of the electrons i.e. a part of rod's kinetic energy. Due to current, opposing force will be produced to slow down the rod i.e. to "oppose the cause" of emf, as per Lenz's law. If the resistance is low, large current will flow and due to greater force, rod will stop quickly. So, more power will be lost in the form of heat but for a short time. If the resistance is high, smaller current will flow, hence, smaller opposing force is produced. Hence less power will be lost in the form of heat but for more time(since rod will take longer time to stop). So, I think the total heat energy lost throughout the motion will be same for a given motional emf. I'm not sure about the last line as I haven't verified it mathematically.

Last edited:
My teacher told that work done by B is zero.

B field is there as a converter. It will convert mechanical energy into electrical energy(E=Blv) and then electrical energy back into mechanical energy(F=Bil). Whatever work is done is at the expense of the kinetic energy given initially to the rod.

theodoros.mihos
Here, electric circuit consumes the kinetic energy of the rod "through" the magnetic field. The opposing mechanical force BiL is the feedback from the circuit to the rod, indicating that the energy is being consumed in the circuit. Hence, the rod slows down. Resistance(conductance, precisely) of the circuit decides how fast the energy is consumed. Its like electrical friction. If friction is more, energy is consumed faster. Like that, if conductance is more, energy is consumed faster and the rod stops in short time. Heat is a part of the kinetic energy of the electrons, just like in any general electrical circuit.

Last edited:
theodoros.mihos

## 1. What exactly is work done by a magnetic field?

Work done by a magnetic field is the amount of energy transferred to a charged particle as it moves through a magnetic field. This can also be thought of as the force exerted on a charged particle multiplied by the distance it is moved.

## 2. How is work done by a magnetic field different from work done by a gravitational field?

The main difference between work done by a magnetic field and work done by a gravitational field is the type of force acting on the charged particle. A magnetic field exerts a force on a charged particle due to its magnetic properties, while a gravitational field exerts a force on a particle due to its mass.

## 3. What are the units of work done by a magnetic field?

The units of work done by a magnetic field are joules (J) or electron volts (eV). This is the same as the units for work done by any other type of force or field.

## 4. Can work done by a magnetic field be positive or negative?

Yes, work done by a magnetic field can have both positive and negative values. If the force and displacement are in the same direction, the work done will be positive. If they are in opposite directions, the work done will be negative.

## 5. How is the work done by a magnetic field calculated?

The work done by a magnetic field is calculated using the equation W = Fd, where W is the work done, F is the force exerted by the magnetic field, and d is the distance the charged particle moves. This equation assumes a constant force and a straight path of motion.

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