# Work done by magnetic field

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1. Dec 24, 2015

### Titan97

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$V_{induced}=Blv$
$F=ilB$

3. 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?

2. Dec 24, 2015

### ElijahRockers

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?

3. Dec 24, 2015

### cnh1995

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: Dec 24, 2015
4. Dec 25, 2015

### Titan97

My teacher told that work done by B is zero.

5. Dec 25, 2015

### cnh1995

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.

6. Dec 25, 2015

### cnh1995

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: Dec 25, 2015