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A circular loop of radius 0.3 cm lies parallel to amuch bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop isA. `9.1 xx 10^(-11)`WbB. `6 xx 10^(-11)`WbC. `3.3 xx 10^(-11)Wb`D. `6.6 xx 10^(-9)`Wb |
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Answer» Correct Answer - A Magnetic flux linked with bigger circular loop is given by `phi = mu_(0)/2.(piIR_(1)^(2)R_(2)^(2))/(R_(1)^(2)+x^(2))^(3//2)` Putting the values, `phi = (4pixx10^(-7)xxpixx15xx(0.3 xx 10^(-2))^(2)xx(20 xx10^(-2))^(2))/(0.3 xx 10^(-2)^(2) + (15 xx 10^(-2))^(2))^(3//2)` By solving, we get, `phi=9.116 xx 10^(-11)`Wb |
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