JEE MAIN - Physics (2006 - No. 9)
A long solenoid has $$200$$ turns per $$cm$$ and carries a current $$i.$$ The magnetic field at its center is $$6.28 \times {10^{ - 2}}\,\,\,Weber/{m^2}.$$ Another long solenoid has $$100$$ turns per $$cm$$ and it carries a current $${i \over 3}$$. The value of the magnetic field at its center is
$$1.05 \times {10^{ - 2}}\,\,Weber/{m^2}$$
$$1.05 \times {10^{ - 5}}\,\,Weber/{m^2}$$
$$1.05 \times {10^{ - 3}}\,\,Weber/{m^2}$$
$$1.05 \times {10^{ - 4}}\,\,Weber/{m^2}$$
Explanation
$${{{B_2}} \over {{B_1}}} = {{{\mu _0}{n_2}{i_2}} \over {{\mu _0}{n_1}{i_1}}}$$
$$ \Rightarrow {{{B_2}} \over {6.28 \times {{10}^{ - 2}}}} = {{100 \times {i \over 3}} \over {200 \times i}}$$
$$ \Rightarrow {B_2} = {{6.28 \times {{10}^{ - 2}}} \over 6}$$
$$ = 1.05 \times {10^{ - 2}}\,\,Wb/{m^2}$$
$$ \Rightarrow {{{B_2}} \over {6.28 \times {{10}^{ - 2}}}} = {{100 \times {i \over 3}} \over {200 \times i}}$$
$$ \Rightarrow {B_2} = {{6.28 \times {{10}^{ - 2}}} \over 6}$$
$$ = 1.05 \times {10^{ - 2}}\,\,Wb/{m^2}$$
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