A straight wire with a resistance of r per unit length is bent to form an angle 2- A rod of the same wire perpendicular to the angle bisector of 2- forms a closed triangular loop- This loop is placed in a uniform magnetic field of a induction B- Calculate the current in the wires when the rod moves at a constant speed V-

Let BC length = l Total length ABC = l+lsinα=l(1+1sinα) Resistance of loop ABC ( R )= rl(1+1sinα) Now,emf induced( e ) acro ...

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Standard XII

Physics

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A straight wi...

Question

A straight wire with a resistance of $r$ per unit length is bent to form an angle $2 α$ . A rod of the same wire perpendicular to the angle bisector (of $2 α)$ forms a closed triangular loop. This loop is placed in a uniform magnetic field of a induction $B$ . Calculate the current in the wires when the rod moves at a constant speed $V$ .

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Solution

Let $B C$ length = $l$
Total length $A B C$ = $l + \frac{l}{sin α} = l (1 + \frac{1}{sin α})$

Resistance of loop $A B C$ ( $R$ )= $r l (1 + \frac{1}{sin α})$
Now,emf induced( $e$ ) across $B C$ , $e = V B l$

Current $i$ = $\frac{e}{R} = \frac{V B l}{r l (1 + \frac{1}{sin α})}$

Therefore, $i = \frac{V B sin α}{r (1 + sin α)}$

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Let BC length =lTotal length ABC =l+lsinα=l(1+1sinα)Resistance of loop ABC(R)=rl(1+1sinα)Now,emf induced(e) across BC , e=VBlCurrent i=eR=VBlrl(1+1sinα)Therefore, i=VBsinαr(1+sinα)