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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^(-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^(-1)` along the negative x-direction (that is it increases by `10^(-3) T cm^(-1 )`as one moves in the negative x-direction), and it is decreasing in time at the rate of `10^(-3) T s^(-1)`. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 `mOmega`. |
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Answer» Side of the square loop, s = 12 cm = 0.12 m Area of the square loop, A =` 0.12 xx 0.12 = 0.0144 m^(2)` Velocity of the loop, v = 8 cm/s = 0.08 m/s Gradient of the magnetic field along negative x-direction, `(dB)/(dx)=10^(-3)Tcm^(-1)=10^(-1)Tm^(-1)` And, rate of decrease of the magnetic field, `(dB)/(dt)=10^(-3)Ts^(-1)` Resistance of the loop, `R=4.5 Omega=4.5xx10^(-3)Omega` Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as: `(dphi)/(dt)=xx(dB)/(dx)xxv` `=144xx10^(-4)m^(2)xx10^(-1)xx0.08` `=11.52xx10^(-5)Tm^(2)s^(-1)` Rate of change of the flux due to explicit time variation in field B is given as: `(dphi)/(dt)=Axx(dB)/(dt)` `=144xx10^(-4)xx10^(-3)` `=1.44xx10^(-5)Tm^(2)s^(-1)` Since the rate of change of the flux is the induced emf, the total induced emf in the loop can be calculated as: `e=1.44xx10^(-5)+11.52xx10^(-5)` `=12.96xx10^(-5)` V Induced current, `i=e/R` `(12.96xx10^(-5))/(4.5xx10^(-3)` `i=2.88xx10^(-2)`A Hence, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction. |
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