JEE MAIN - Physics (2023 - 13th April Evening Shift - No. 22)
A straight wire $$\mathrm{AB}$$ of mass $$40 \mathrm{~g}$$ and length $$50 \mathrm{~cm}$$ is suspended by a pair of flexible leads in uniform magnetic field of magnitude $$0.40 \mathrm{~T}$$ as shown in the figure. The magnitude of the current required in the wire to remove the tension in the supporting leads is ___________ A.
$$\left(\right.$$ Take $$g=10 \mathrm{~ms}^{-2}$$ ).
_13th_April_Evening_Shift_en_22_1.png)
Answer
2
Explanation
In the given situation, the wire is suspended by a pair of flexible leads in a uniform magnetic field. Due to the magnetic field, the wire experiences a magnetic force which causes it to hang at an angle to the vertical. The tension in the flexible leads provides the restoring force to balance the weight of the wire.
For equilibrium, the magnetic force on the wire should balance the weight of the wire. Therefore, we can write:
$$ \mathrm{Mg}=\mathrm{I} \ell \mathrm{B} $$
where $\mathrm{M}$ is the magnetic force on the wire, $\mathrm{g}$ is the acceleration due to gravity, $\mathrm{I}$ is the current flowing through the wire, $\ell$ is the length of the wire, and $\mathrm{B}$ is the magnitude of the magnetic field.
Solving for $\mathrm{I}$, we get:
$$ \mathrm{I}=\frac{\mathrm{mg}}{\ell \mathrm{B}} $$
Substituting the given values, we get:
$$ \mathrm{I}=\frac{40 \times 10^{-3} \times 10}{50 \times 10^{-2} \times 0.4}=2 \mathrm{~A} $$
Therefore, the magnitude of the current required in the wire to remove the tension in the supporting leads is 2 A.
For equilibrium, the magnetic force on the wire should balance the weight of the wire. Therefore, we can write:
$$ \mathrm{Mg}=\mathrm{I} \ell \mathrm{B} $$
where $\mathrm{M}$ is the magnetic force on the wire, $\mathrm{g}$ is the acceleration due to gravity, $\mathrm{I}$ is the current flowing through the wire, $\ell$ is the length of the wire, and $\mathrm{B}$ is the magnitude of the magnetic field.
Solving for $\mathrm{I}$, we get:
$$ \mathrm{I}=\frac{\mathrm{mg}}{\ell \mathrm{B}} $$
Substituting the given values, we get:
$$ \mathrm{I}=\frac{40 \times 10^{-3} \times 10}{50 \times 10^{-2} \times 0.4}=2 \mathrm{~A} $$
Therefore, the magnitude of the current required in the wire to remove the tension in the supporting leads is 2 A.
Comments (0)
