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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» Solution :Here side of square loop `l = 12 cm = 0.12 m`, hence area of square loop `A = l^(2) = (0.12)^(2) = 1.44 xx 10^(-2)m^(2)`, Velocity of loop `V = 8cm s^(-1) = 0.08 m s^(-1)` along + ve x-direction, Along + vex-direction field gradient `(dB)/(dx)=-10^(-3)Tcm^(-1)=-10^(-1)Tm^(-1),` and rate of CHANGE of field with time `(db)/(dx) = - 10^(-3) T s^(-1)` `therefore` Induced cmf due to change in field with distance `varepsilon=-d/dt(BA)=-A(dB)/(dt)=-A(dB)/dx.dx/dt=-A(dB)/dxv` `=-(1.44xx10^(-2))xx(-10^(-1))xx(0.08)=11.52xx10^(-5)V` and induced emf due to change in field with time `varepsilon_(1)=-A(dB)/dt=-1.44xx10^(-2)xx(-10^(-3))=1.44xx10^(-5)V` `therefore` Total induced emf `varepsilon = varepsilon_(1)+ varepsilon_(2)=11.52 xx10^(-5)+1.44xx10^(-5)=12.96xx10^(-5)V` As resistance of loop `R = 4.5 mOmega = 4.5 xx 10^(-3) OMEGA` `therefore` Induced current `I=varepsilon/R=(12.96xx10^(-5))/(4.5xx10^(3))=2.88xx10^(-2)A`. As the magnetic field is along +ve z direction and motion of the loop is along the +ve x-direction, along which magnetic field is decreasing. Hence, on applying Lenz.s law we find that for an observer, who observes the loop to be moving towards RIGHT, the induced current will be seen to be flowing anticlockwise. |
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