First slide

Applications involving charged particles moving in a magnetic field

Question

A wire of cross-sectional area A forms three sides of a square and is free to rotate about axis $O O^{1}$ . If the structure is deflected by an angle $θ$ from the vertical when current $i$ is passed through it in a magnetic field B acting vertically upwards and density of the wire is $ρ$ , then the value of $θ$ is given by

Difficult

A

$\frac{2 A ρ g}{i B} = \cot θ$
B

$\frac{2 A ρ g}{i B} = \tan θ$
C

$\frac{A ρ g}{i B} = \sin θ$
D

$\frac{A ρ g}{2 i B} = \cos θ$

Solution

Let a is the side of square.
Torque of current = $M B = \sin (90 - θ) = M B \cos θ = i a^{2} B \cos θ$
Mass of each side= $m = ρ A a$
Torque of gravity = $2 m g \frac{a}{2} \sin θ + m g a \sin θ$ $= 2 m g a \sin θ = 2 ρ A g a^{2} \sin θ$
Now, at equilibrium both torques should be same
i.e., $i a^{2} B \cos θ = 2 ρ A g a^{2} \sin θ$
$\Rightarrow \cot θ = \frac{2 ρ A g}{i B}$ .

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