Question
An equilateral triangular loop $$ABC$$ made of uniform thin wires is being pulled out of a region with a uniform speed $$v,$$ where a uniform magnetic field $${\vec B}$$ perpendicular to the plane of the loop exists. At time $$t = 0,$$ the point $$A$$ is at the edge of the magnetic field. The induced current $$\left( I \right)$$ vs time $$\left( t \right)$$ graph will be as
An equilateral triangular loop $$ABC$$ made of uniform thin wires is being pulled out of a region with a uniform speed $$v,$$ where a uniform magnetic field $${\vec B}$$ perpendicular to the plane of the loop exists. At time $$t = 0,$$ the point $$A$$ is at the edge of the magnetic field. The induced current $$\left( I \right)$$ vs time $$\left( t \right)$$ graph will be as
A.
B.
C.
D.
Answer :
Solution :
$$\eqalign{ & I = \frac{\varepsilon }{R} = \frac{{\left| {\frac{{d\phi }}{{dt}}} \right|}}{R} = \frac{{B.\frac{{dA}}{{dt}}}}{R} \cr & = \frac{{B\frac{d}{{dt}}\left( {\frac{1}{2}h.b} \right)}}{R} \propto \frac{{Bb\frac{{dh}}{{dt}}}}{R} \propto Bbv \cr & \therefore b \propto t \Rightarrow I \propto t \cr} $$
$$\eqalign{ & I = \frac{\varepsilon }{R} = \frac{{\left| {\frac{{d\phi }}{{dt}}} \right|}}{R} = \frac{{B.\frac{{dA}}{{dt}}}}{R} \cr & = \frac{{B\frac{d}{{dt}}\left( {\frac{1}{2}h.b} \right)}}{R} \propto \frac{{Bb\frac{{dh}}{{dt}}}}{R} \propto Bbv \cr & \therefore b \propto t \Rightarrow I \propto t \cr} $$
