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A wire carrying current I is tied between points P and Q and is in the shape of a circular arch of radius R due to a uniform magnetic field B (perpendicular to the plane of the paper, shown by xxx) in the vicinity of the wire. If the wire subtends an angle $2\Theta _{0}$ at the centre of the circle (of which it forms an arch) then the tension in the wire is :

Option 1)
IBR

Option 2)
$\frac{IBR}{\sin \Theta _{0}}$

Option 3)
$\frac{IBR}{2\sin \Theta _{0}}$

Option 4)
$\frac{IBR\: \Theta_{0} }{\sin \Theta _{0}}$

Answers (1)

best_answer

As we learnt in

Total magnetic force -

$\underset{F}{\rightarrow}=i ( \underset{L}{\rightarrow}\times \underset{B}{\rightarrow})$

- wherein

Magnetic force on the

Total magnetic force -

$\underset{F}{\rightarrow}=i ( \underset{L}{\rightarrow}\times \underset{B}{\rightarrow})$

- wherein

Magnetic force on the circular arc $F=I \left ( 2R sin\theta \right )\pi$

where $2T sin\theta =F$

$2T sin\theta = 2\pi R sinB\: \: \: \: \: \: \: \: \therefore T=IRB$

Option 1)

IBR

Correct Option

Option 2)

$\frac{IBR}{\sin \Theta _{0}}$

Incorrect Option

Option 3)

$\frac{IBR}{2\sin \Theta _{0}}$

Incorrect Option

Option 4)

$\frac{IBR\: \Theta_{0} }{\sin \Theta _{0}}$

Incorrect Option

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