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The earth's magnetic field at a certain point is \( 7.0 \times \) \( 10^{-5} T \). This field is to be balanced by a magnetic field at the centre of a circular current carrying coil of radius \( 5 cm \) by suitably orienting it. If the coil has 100 turns then the required current (up to nearest integer) is \( X \) \( mA \). Find \( X \). |
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Answer» Magnetic induction due to circular coil at centre \(B =\frac{\mu_0NI}{2R}\) Then \(I = \frac{B.2R}{\mu_0N}\) \(I = \frac{7\times 10^{-5}\times 2\times 0.05}{4\pi \times 10^{-7}\times 100}\) \(I = \frac{7\times 10^{-5}\times 2\times 0.05}{4\times 3.14 \times 10^{-7} \times 100}\) \(I = \frac{0.7\times 10^{-5}}{12.56\times 10^{-5}}\) \(I = 0.055\) \(I = 55\times 10^{-3} A\) \(I = 55 mA\) Then \(X = 55\) |
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