Two ends of a current carrying wire AB of length l, radius r and resistivity of material ρ are connected to the terminals of a battery of emf E. The wire AB is placed in a uniform magnetic field of induction B→ as shown. Force experienced by the wire is 20 N. It radius of the wire is doubled and length also doubled, force experienced by the wire will be

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10 N

40 N

80 N

60 N

answer is C.

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Detailed Solution

Current through the wire is given by i=ER=Eρ.l/πr2=E.πr2ρ.l⇒i∝r2lF=Bil∴F ∝r2 F1F=r1r2⇒F1=4×20 N=80 N

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