JEE MAIN - Physics (2008 - No. 8)
A horizontal overhead powerline is at height of $$4m$$ from the ground and carries a current of $$100A$$ from east to west. The magnetic field directly below it on the ground is
$$\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,\,Tm\,\,{A^{ - 1}}} \right)$$
$$\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,\,Tm\,\,{A^{ - 1}}} \right)$$
$$2.5 \times {10^{ - 7}}\,T$$ southward
$$5 \times {10^{ - 6}}\,T$$ northward
$$5 \times {10^{ - 6}}\,T$$ southward
$$2.5 \times {10^{ - 7}}\,T$$ northward
Explanation
The magnetic field is
$$B = {{{\mu _0}} \over {4\pi }}{{2I} \over r}$$
$$ = {10^{ - 7}} \times {{2 \times 100} \over 4}$$
$$ = 5 \times {10^{ - 6}}T$$
According to right hand palm rule, the magnetic field is directed towards south.
$$B = {{{\mu _0}} \over {4\pi }}{{2I} \over r}$$
$$ = {10^{ - 7}} \times {{2 \times 100} \over 4}$$
$$ = 5 \times {10^{ - 6}}T$$
According to right hand palm rule, the magnetic field is directed towards south.
Comments (0)
