Question

A long metal bar of 30 cm length is aligned along a north south line and moves eastward at a speed of 10 m/s. A uniform magnetic field of 4.0 T point vertically downwards. If the south end of the bar has a potential of 0 V, the induced potential at the north end of the bar is:

A. \[ + 12\,{\text{V}}\]
B. \[ - 12\,{\text{V}}\]
C. 0 V
D. Cannot be determined since there is no closed circuit.

Answer 1

Hint:When we move the metal bar in the uniform magnetic field with some velocity, an emf induced in the metal bar. Use the right-hand rule to determine the direction of this induced emf. Use the formula for induced emf in the metal bar moving in the uniform magnetic field.

Formula used:
Induced emf, \[\varepsilon = Blv\]
Here, B is the magnetic field, l is the length of the metal bar and v is the velocity of the metal bar.

Complete step by step answer:
We have given the length of metal bar \[l = 30\,{\text{cm}} = 0.3\,{\text{m}}\], the magnitude of magnetic field \[B = 4.0\,{\text{T}}\] and velocity of the metal bar is \[v = 10\,{\text{m/s}}\]. We know that when we move the metal bar in the uniform magnetic field with some velocity, an emf induced in the metal bar. We can determine the direction of the emf using the right-hand rule.

According to right-hand rule if we hold the fingers of our right hand perpendicular to each other such that the thumb represents the direction of the magnetic field, the middle finger represents the direction of motion of the metal bar then the direction of the forefinger represents the direction of the induced emf. Thus, the direction of the induced emf is along the northward direction. We have the expression for the induced emf in the moving metal bar in the uniform magnetic field,
\[\varepsilon = Blv\]
Here, B is the magnetic field, l is the length of the metal bar and v is the velocity of the metal bar.
Substituting \[B = 4.0\,{\text{T}}\], \[l = 0.3\,{\text{m}}\] and \[v = 10\,{\text{m/s}}\] in the above equation, we get,
\[\varepsilon = \left( 4 \right)\left( {0.3} \right)\left( {10} \right)\]
\[ \therefore \varepsilon = 12\,{\text{V}}\]
Therefore, since the south end of the metal bar has potential 0 V, the north end will have the potential of 12 V.

So, the correct answer is option A.

Note:The induced emf will be zero only when the metal bar is not moving. To use the formula, \[\varepsilon = Blv\], the unit of the magnetic field must be tesla. To determine the direction of the induced emf, students must use right-hand rule and not right-hand thumb rule or Fleming’s right-hand rule.