Magnetic Effects & Magnetism

JEE Main level, single correct, medium

205 of 276

Question 205 of 276

Two long straight wires PP and QQ carrying equal current 10A10A each were kept parallel to each other at 5cm5cm distance. Magnitude of magnetic force experienced by 10cm10cm length of wire PP is F1{F}_{1}. If distance between wires is halved and currents on them are doubled, force F2{F}_{2} on 10cm10cm length of wire PP will be:

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Answer: (A) 8F18{F}_{1}

As we know, force per unit length between two parallel straight wires is given by, F=μ0i1i22πdF=\dfrac{{\mu }_{0}{i}_{1}{i}_{2}}{2\pi d}.
Therefore, required ratio
F1F2=μ0(10)22π(5×10−2)μ0(20)22π(52×10−2)=18\dfrac{{F}_{1}}{{F}_{2}}=\dfrac{\dfrac{{\mu }_{0}(10{)}^{2}}{2\pi (5\times {10}^{-2})}}{\dfrac{{\mu }_{0}(20{)}^{2}}{2\pi \left(\dfrac{5}{2}\times {10}^{-2}\right)}}=\dfrac{1}{8}
⇒F2=8F1\Rightarrow {F}_{2}=8{F}_{1}

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