JEE MAIN - Physics (2020 - 7th January Morning Slot - No. 20)
A loop ABCDEFA of straight edges has six corner points A(0, 0, 0), B(5, 0, 0), C(5, 5, 0), D (0, 5,
0), E(0, 5, 5) and F(0, 0, 5). The magnetic field in this region is $$\overrightarrow B = \left( {3\widehat i + 4\widehat k} \right)T$$
. The quantity of
flux through the loop ABCDEFA (in Wb) is _______.
Risposta
175
Spiegazione
$$\phi $$ = $$\overrightarrow B .\overrightarrow A $$ = $$\left( {3\widehat i + 4\widehat k} \right).\left( {25\widehat i + 25\widehat k} \right)$$
$$ \Rightarrow $$ $$\phi $$ = (3 $$ \times $$ 25) + (4 $$ \times $$ 25) = 175 weber
$$ \Rightarrow $$ $$\phi $$ = (3 $$ \times $$ 25) + (4 $$ \times $$ 25) = 175 weber