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NCERT Class XII Chapter
Electromagnetic Induction
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Question : 12 of 17
Marks: +1, -0
A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s−1s^{-1} in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10−3 T cm−110^{-3}\,\mathrm{T}\,\mathrm{cm}^{-1} along the negative x-direction (that is it increases by 10−3 T cm−110^{-3}\,\mathrm{T}\,\mathrm{cm}^{-1} as one moves in the negative x-direction) and it is decreasing in time at the rate of 10–3 T s−1s^{-1}. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ
Solution:  
Each side of square loop is 12 cm and magnetic field is decreasing along x direction.
dBdx\frac{dB}{dx} = - 10−3Tcm−110^{-3}\mathrm{T}\mathrm{cm}^{-1} = - 0.1 T m−1m^{-1}
Also the magnetic field is decreasing with time at constant rate
dBdt\frac{dB}{dt} = - 10−3Ts−110^{-3}\mathrm{T}\mathrm{s}^{-1}
Induced emf and rate of change of magnetic flux due to only time variation
εt\varepsilon_t = - dϕdt\frac{d\phi}{dt} = - dBAdt\frac{dBA}{dt} = - A dBdt\frac{dB}{dt} or εt\varepsilon_t = - 0.12 × 0.12 [−10−3][-10^{-3}] = 144 × 10−710^{-7} V
Induced emf and rate of change of magnetic flux due to change in position.
εx\varepsilon_x = - dBAdt\frac{dBA}{dt} = - A dBdx\frac{dB}{dx} × dxdt\frac{dx}{dt}
εx\varepsilon_x = - Av dBdx\frac{dB}{dx} = -0.12 × 0.12 × 0.08 × (0.1) = 1152 × 10−710^{-7} V
Both the induced emf have same sign and thus adds to provide net Induced emf in the loop
εnet\varepsilon_{\text{net}} = εi+εx\varepsilon_i+\varepsilon_x = 1296 × 10−710^{-7} V
Induced current
I = εnetR\frac{\varepsilon_{\text{net}}}{R} = 1296×10−74.5×10−3\frac{1296\times10^{-7}}{4.5\times10^{-3}} = 2.88 × 10−210^{-2} A
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