Saved Bookmarks
| 1. |
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 . |
Answer» Solution :Here magnetic field changes with position and time both. Hence we have two values of induced emf say `epsilon_1` and `epsilon_2` respectively. Situation given in the STATEMENT is shown in the figure given below.  `epsilon_1=-(dphi)/(dt)=-d/(dt)(AB cos 0^@) (because vecA||vecB)` `therefore epsilon_1=-A (dB)/(dt)` `=-A("dB"/"dx")("dx"/"dt")` `=-l^2(dB)/(dx)v` `=-(0.12)^2 (-10^(-3)/10^(-2))(8xx10^(-2))` ( `because (dB)/(dx)` along +X axis is `-10^(-3) T/"CM"`) `therefore epsilon_1=1.152xx10^(-4)` V....(1) `epsilon_2=-(dphi)/(dt)=-d/(dt)(AB)` `therefore epsilon_2=-A (dB)/(dt)` `=-l^2/(dB)/(dt)` `=-(0.12)^2(-10^(-3))` `therefore epsilon_2=0.144xx10^(-4)` V ...(2) Induced current `I=("effective induced emf"(epsilon))/"equivalent resistance (R)"` `therefore I=(epsilon_1+epsilon_2)/R` ( `because` Here `epsilon_1 gt 0, epsilon_2 gt 0 rArr epsilon=epsilon_1+epsilon_2`) `therefore I=((1.152+0.144)xx10^(-4))/(4.5xx10^(-3))` `therefore I=2.88xx10^(-2)`A Here, as the loop moves along + X axis, outward magnetic flux linked with it GOES on decreasing and so according to Lenz.s law, induced current should flow anticlockwise in the loop so that outward magnetic flux can increase. |
|