Question

A wire of length \[50{\rm{ }}cm\] moves with a velocity of\[300{\rm{ }}m/min\], perpendicular to a magnetic field. If the emf induced in the wire is\[2{\rm{ }}V\], the magnitude of the field in tesla is
1. \[2\]
2. \[5\]
3. \[0.8\]
4. \[2.5\]

Answer 1

Hint: here, in this question it has been asked that the wire moves with velocity and that wire is perpendicular to the magnetic field. Emf induced is given and we have to find out the magnitude of the magnetic field. For an easier view just draw a diagram explaining the above question. We have to use the relation between emf and magnetic field for obtaining the answer.

Formula used:
\[e = Blv\]

Complete answer:
Let us consider a figure drawn below for more illustrative view of the given question,

Given data,
length of the wire, \[l = 50cm\]
Velocity of the wire moving in the magnetic field, \[v = 300m/\min \]
Emf induced in the wire, \[e = 2{\rm{ }}V\]

Thus, by using relation between emf and magnetic field \[B\]. Therefore,
\[e = Blv\]

Implementing the given values in the above formula, we get
\[2{\rm{ }}V = B(50cm)(300m/\min )\]

Here, velocity is given in m/min we have to convert it into m/s, since it is standard unit of measurement of velocity.
\[\therefore 2{\rm{ }}V = B(0.5m)\left( {\frac{{300m}}{{60s}}} \right)\]
\[\therefore B = 0.8T\]

After calculating we get magnetic field as \[0.8T\]. Answer: \[0.8T\], Option: 3.

Note: While solving such questions, always make sure you change all the needed and used physical quantities and numerical values in the same set of standard units and if a wire isn't moving perpendicularly then only the perpendicular component of velocity of the wire is used to calculate emf on the wire.