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A straight wire of mass $200\;{\text{g}}$ and length $1.5\;{\text{m}}$ carries a current of $2\;{\text{A}}$. It is suspended in mid-air by a uniform horizontal magnetic field ${\text{B}}$. The magnitude of ${\text{B}}$ (in tesla) is:
(A) \[0.65\]
(B) \[0.55\]
(C) \[0.75\]
(D) \[0.45\]

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Last updated date: 03rd Mar 2024
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Answer
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Hint: As we can all see, the force acting under a uniform magnetic field on a current-carrying wire is equal to the weight of the wire because the wire is in equilibrium and has no acceleration, so it does not move.
Formula Used: We will be using the following formula,
\[F = BiL\sin \theta \]
Where
\[F\] is the force due to the magnetic field
\[B\] is the magnitude of the magnetic field
\[i\] is the current carrying in the wire
\[L\] is the length of the wire
\[\theta \] in the angle between magnetic field and current

Complete Step-by-Step solution:
The following information is provided to us in the given question:
The mass of the wire, \[m = 200g = 0.2kg\]
The length of the wire, \[l = 15m\]
The current flowing through the wire, \[i = 2A\]
And we have to find out the magnetic field
Let us draw the stick figure of the wire given in the question and mention all the forces acting on it
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Now, we can see that the direction of current is perpendicular to the magnetic field horizontally, hence the angle
$\theta = {90^\circ }$
We know that
\[F = BiL\sin \theta \]
Since, $\theta = {90^\circ }$, we have \[\sin {90^ \circ } = 1\]
So,
\[F = BiL\]
Now the weight of the wire is given by \[w = mg\]
That is \[w = 0.2 \times 9.8 N\]
According to the figure,
\[F = mg\]
Now, let us substitute the known values in the formula to get
\[BiL = mg\]
Rearranging the terms, we have
\[B = \dfrac{{mg}}{{iL}}\]
Let us now substitute the values
\[B = \dfrac{{0.2 \times 9.8}}{{2 \times 15}}\]
So, we get
\[\therefore B = 0.65 T\]

Hence, the correct option is (A.)

Note: We all learned that the left-hand rule of Fleming is used to describe the force placed at right angles to a magnetic field on a current-carrying conductor. The conductor carrying the current should be made of ferromagnetic material.
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