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Figure given in the question is a cross-sectional view of a coaxial cable. The centre conductor. surrounded by a rubber layer, which is surrounded by an outer conductor, which is surround by another rubber layer. The current in the inner conductor is 1.0 A out of the page, and the current in the outer conductor is 3.0 A into the page. Determine the magnitude and direction of the magnetic field at points a and b. |
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Answer» Solution :Applying Ampere's circular circuital law. `B_A=mu_0/(2PI) i_("in")/r` `=((2xx10^-7)(1))/10^-3` `=2xx10^-4T` This due to `(*)` current of `1A`. HENCE, MAGNETIC lines are circular and anti-clockwise. Hence, magnetic field is upwards. `B_b=mu_0/(2pi) i_("in")/r` `=((2xx10^-7)(3-1))/(3xx10^-3)` `=1.33xx10^-4T` This is due to net `(ox)` current. Hence, magnetic lines are clockwise. So, magnetic fied at `B` is downwards. |
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