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A coil carrying a current I=10 ~mA is placed in a uniform magnetic field so that its axis coincides with the field direction. The coil consist of only one turn and made up of copper. The diameter of the wire is 0.1 mm , the radius of the coil is R=3 cm . The value of external induction field will the coil rupture is 500 K . Breaking stress of copper =3 10^8 ~N / m ^2 . Then find the value of K .

Correct answer

5

Step-by-step solution

aligned & 2 T 2 = Bi ( R ) & T = IRB & also, T= r ^2 & B = r ^2 IR aligned

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