Question

A metallic wire having length of 2 m and weight of $4\times {{10}^{-3}}N$ is found to remain at rest in a uniform and transverse magnetic field $2\times {{10}^{-4}}T$. Current flowing through the wire is
A. 10A
B. 5A
C. 2A
D. 1A

Answer 1

Hint: Study about the magnetic force on a current carrying wire placed in a magnetic field. The magnetic force is given by the product of the current flowing through the wire, length of the wire and the magnitude of the magnetic field the wire is placed in. since the wire is at rest try to equate the magnetic force and the weight of the wire. Then we can find the answer.

Complete Step-by-Step solution:
We have a wire of length, $l=2m$
The weight of wire is, $W=4\times {{10}^{-3}}N$
The wire is at rest in a uniform and transverse magnetic field, $B=2\times {{10}^{-4}}T$
Again, the wire remains at rest. So, the force on the wire is zero. That means the weight of the wire is equal to the magnetic force on the wire.
Now, magnetic force on a wire in a magnetic field is given by,
${{F}_{m}}=IlB$
Where I is the current flowing through the wire, l is the length of wire and B is the magnetic field in which the wire is placed.
Now, we equate the weight of the wire to magnetic force applied on it and get,
$\begin{align}
  & {{F}_{M}}=W \\
 & IlB=W \\
 & I=\dfrac{W}{lB} \\
\end{align}$
Putting the values of the given quantities in the above equation we can get,
$\begin{align}
  & I=\dfrac{4\times {{10}^{-3}}}{2\times 2\times {{10}^{-4}}} \\
 & I=10A \\
\end{align}$
So, the current flowing through the wire is 10 A.
The correct option is (A).

Note: The direction of the magnetic force on a current carrying wire can b e given by the right-hand thumb rule. If we curl our fingers from the direction of current towards the direction of the magnetic field, the direction of the magnetic force is given by the direction of the thumb. The magnetic force is perpendicular to both the wire and the magnetic field.