Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

213 of 276

Question 213 of 276

A coil in the shape of an equilateral triangle of side 10cm10cm lies in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field 20mT20mT. The torque acting on the coil when a current of 0.2A0.2A is passed through it and its plane becomes parallel to the magnetic field will be x×10−5Nm\sqrt{x}\times {10}^{-5}Nm. The value of xx is

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Answer: 3

τ=M⃗×B⃗\tau =\vec{M}\times \vec{B}
=IAB=IAB
=(0.2)×(3a24)×(2×10−2)=0.2×3×10−2×2×10−24=\left(0.2\right)\times \left(\dfrac{\sqrt{3}{a}^{2}}{4}\right)\times \left(2\times {10}^{-2}\right)=\dfrac{0.2\times \sqrt{3}\times {10}^{-2}\times 2\times {10}^{-2}}{4}
=3×10−5Nm=\sqrt{3}\times {10}^{-5}Nm

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