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A magnetic field induction is changing in magnitude in a region at a constant rate `dB//dt`. A given mass `m` of copper drawn into a wire and formed into a loop is placed perpendicular to the field. If the values of specific resistance and density of copper are `rho` and `sigma` respectively, then the current in the loop is given by :A. (a) `(m)/(2pirhod)(dB)/(dt)`B. (b) `(m)/(4pia^(2)r)(dB)/(dt)`C. (c) `(m)/(4piad)(dB)/(dt)`D. (d) `(m)/(4pirhod)(dB)/(dt)` |
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Answer» Correct Answer - D (d) `xi = (dphi)/(dt) `rarr` I = (xi)/(R )(d)/(dA)(BA) = (A)/(R )(dB)/(dt)` where `pir^(2)` = area of the loop of radius `r` and `R` = resistance of the loop of length `(2pir)` and area of cross section `pia^(2)`. `R = (rhol)/(pia^(2)) = (rho(2pir))/(pia^(2))` Further mass of wire is `m = (pia^(2))(2pir)(d)` `i = ((pia^(2))(pir^(2)))/(rho(2pir))(dB)/(dt)` `i = ((pia^(2))(pir^(2)))/(4pirho)(dB)/(dt) rarr i = (m)/(4pirhod)(dB)/(dt)` |
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