JEE MAIN - Physics (2020 - 4th September Morning Slot - No. 13)
A wire A, bent in the shape of an arc of a circle, carrying a current of 2 A and having radius 2 cm and another wire B, also bent in the shape of arc of a circle, carrying a current of 3 A and having radius of 4 cm, are placed as shown in the figure. The ratio of the magnetic fields due to the wires A and B at the common centre O is :
_4th_September_Morning_Slot_en_13_1.png)
4 : 6
6 : 4
2 : 5
6 : 5
Explanation
Using formula, $$B = {{{\mu _0}i} \over {2r}}\left( {{\theta \over {2\pi }}} \right)$$
$${B_A} = {{\mu (2)\left( {{{3\pi } \over 2}} \right)} \over {2(a)(2\pi )}} = {{3\mu } \over {4a}}$$
$${B_B} = {{\mu (3)\left( {{{5\pi } \over 3}} \right)} \over {2(2a)(2\pi )}} = {{5\mu } \over {8a}}$$
$$ \therefore $$ $${{{B_A}} \over {{B_B}}} = {{3\mu } \over {4a}} \times {{8a} \over {5\mu }} = 6:5$$
$${B_A} = {{\mu (2)\left( {{{3\pi } \over 2}} \right)} \over {2(a)(2\pi )}} = {{3\mu } \over {4a}}$$
$${B_B} = {{\mu (3)\left( {{{5\pi } \over 3}} \right)} \over {2(2a)(2\pi )}} = {{5\mu } \over {8a}}$$
$$ \therefore $$ $${{{B_A}} \over {{B_B}}} = {{3\mu } \over {4a}} \times {{8a} \over {5\mu }} = 6:5$$
Comments (0)
