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A rectangular loop of area 20xx30cm is placed in a magnetic field of 0.3T with its plane (i) normal to the field (ii) inclined 30^(@) to the field and (iii) parallel to the field. Find the flux linked with the coil in each case. |
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Answer» Solution :Given : `A=20cmxx30cm` B = 0.3T Let `theta` be the angle made by the field B with the normal to the PLANE of the coil. i. Here, `theta=90^(@)-90^(@)=0^(2)` So,flux `phi=BAcostheta` `=0.3xx6xx10^(-2)xxcos0^(@)` `e=1.8xx10^(-2)Wb` II. Here, `theta=90^(@)-30^(@)=60^(@)` `phi=0.3xx6xx10^(-2)xxcos60^(@)` `phi=0.9xx10^(-2)Wb` iii. Here, `theta=90^(@)-0^(@)=90^(@)` `phi=0.3xx6xx10^(-2)xxcos90^(@)` `phi=0^(@)` |
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