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)

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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