A straight piece of conducting wire mass M and length L is m

A straight piece of conducting wire, mass M and length L, is moved through a region of uniform magnetic field B. as shown. Which end will be positive and which end will be negative? Explain. Which end is at higher potential? Explain. How strong is the magnetic field if the potential difference between the ends of the rod is 1.50 V. Assume you are told the rod\'s length is 2.50 m and it is being moved with a speed of 300 m/s (which is just under the speed of sound). Show clearly the direction of the induced current in each of the following listed objects, in the given situation shown below

Solution

For a rod of length L, being moved with velocity v in a ,magnetic field B, the EMF induced is given as E = BLv

Now it needs to be noted here that the induction happens as a result of motion of hte electrons within the rod which in turn are being moved with velocity v across the magnetic field. The force acting on the electrons is given as

F = qV x B where the direction of the force is along the cross product of V and B for q being positive and is opposite to that for q being negative.

Here the V x B will be along the right hand end of the rod shown above, so the electrons will be pushed away. Therefore the right end (The one near which I see \'M\' written) will be at higher potential and will be positive, while the other end will be negative hence at lower potential.

We have been given that the potential difference is 1.5 V and the length of the rod is 2.5 m with v = 300 m/s

So, we can write: E = B x 2.5 x 300 = 1.5

or, B = 2 x 10^-3 T is the required magnetic field.