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Question

# A coil in the shape of an equilateral triangle of side 0.02 m is suspended from a vertex such that it is hanging in a vertical plane. A magnetic field of magnetic strength 5×10−2 T is applied along the plane of coil. When a current of 0.1 A is passed through the coil, the torque experienced by its will be:

A
53×107 Nm
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B
22×107 Nm
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C
105×107 Nm
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D
5×105 Nm
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Solution

## The correct option is A 5√3×10−7 Nm The diagram shows the arrangement of coil & magnetic field. As the coil is in the form of equilateral triangle, its area is, A=12×L×Lsin600=12×(0.02)2×√32 ⇒A=√3×10−4 m2 Now, the magnetic moment of the coil: μ=iA=0.1×√3×10−4=√3×10−5 Am2 The torque acting on the coil will be, →τ=→μ×→B ⇒τ=μBsinθ Since, the plane of the coil is parallel to the magnetic field, the vector(→A) will be perpendicular to the magnetic field. Thus, →μ⊥→B and θ=900 ⇒τ=μB=√3×10−5×5×10−2 ∴τ=5√3×10−7 Nm Hence, option (a) is the correct answer.

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