WBJEE · Physics · Magnetic Effects of Current
A straight wire of length \(2 \mathrm{m}\) carries a current of \(10 \mathrm{A}\). If this wire is placed in a uniform magnetic field of 0.15 T making an angle of \(45^{\circ}\) with the magnetic field, the applied force on the wire will be
- A \(1.5 \mathrm{N}\)
- B \(3 \mathrm{N}\)
- C \(3 \sqrt{2} \mathrm{N}\)
- D \(\frac{3}{\sqrt{2}} \mathrm{N}\)
Answer & Solution
Correct Answer
(D) \(\frac{3}{\sqrt{2}} \mathrm{N}\)
Step-by-step Solution
Detailed explanation
Given \(i=10 \mathrm{A}, B=0.15 \mathrm{T}, \theta=45^{\circ}\) and \(l=2 \mathrm{m}\) Here, \[ \begin{aligned} F &=i L B \sin \theta \\ &=10 \times 2 \times 0.15 ~ x ~ \sin 45^{\circ} \\ &=\frac{3}{\sqrt{2}} \mathrm{N} \end{aligned} \]
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