MHT CET · Physics · Magnetic Effects of Current
Two wires \(2 \mathrm{~mm}\) apart supply current to a \(100 \mathrm{~V}\), \(1 \mathrm{~kW}\) heater. The force per metre between the wires is \(\left(\mu_0=4 \pi \times 10^{-27}\right.\) SI unit)
- A \(2 \times 10^{-2} \mathrm{~N}\)
- B \(4 \times 10^{-3} \mathrm{~N}\)
- C \(2 \times 10^2 \mathrm{~N}\)
- D \(10^{-2} \mathrm{~N}\)
Answer & Solution
Correct Answer
(D) \(10^{-2} \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
Given: \(P=1 \mathrm{~W}=1000 \mathrm{~W}, \mathrm{~V}=100 \mathrm{~V}\)
\(\therefore\mathrm{I}=10 \mathrm{~A} \quad \ldots .(\because \mathrm{P}=\mathrm{VI})\)
\(\therefore \mathrm{F}=\frac{\mu_0 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \times \mathrm{a}}=\frac{4 \pi \times 10^{-7} \times 100}{2 \pi \times\left(2 \times 10^{-3}\right)}=10^{-7} \times 10^5=10^{-2} \mathrm{~N}\)
\(\therefore\mathrm{I}=10 \mathrm{~A} \quad \ldots .(\because \mathrm{P}=\mathrm{VI})\)
\(\therefore \mathrm{F}=\frac{\mu_0 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \times \mathrm{a}}=\frac{4 \pi \times 10^{-7} \times 100}{2 \pi \times\left(2 \times 10^{-3}\right)}=10^{-7} \times 10^5=10^{-2} \mathrm{~N}\)
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