JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A coil in the shape of an equilateral triangle of side \(10\, {cm}\) lies in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field \(20\, {mT}\). The torque acting on the coil when a current of \(0.2\, {A}\) is passed through it and its plane becomes parallel to the magnetic field will be \(\sqrt{{x}} \times 10^{-5} \,{Nm}\). The value of \({x}\) is ..... .
- A \(3\)
- B \(7\)
- C \(10\)
- D \(13\)
Answer & Solution
Correct Answer
(A) \(3\)
Step-by-step Solution
Detailed explanation
\(\vec{\tau}=\vec{M} \times \vec{B}=M B \sin 90^{\circ}\) \(=M B=\frac{i \sqrt{3} \ell^{2}}{4} B\) \(=\sqrt{3} \times 10^{-5} {N}-{m}\)
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