The mean square current is
\(\bar { { I }^{ 2 } } =\frac { 1 }{ \tau } \int _{ 0 }^{ \tau }{ { I }_{ 0 }^{ 2 } } \left( t/\tau \right) dt=\frac { { I }_{ 0 }^{ 2 } }{ { \tau }^{ 3 } } \int _{ 0 }^{ \tau }{ { t }^{ 2 } } dt=\frac { { I }_{ 0 }^{ 2 } }{ { \tau }^{ 3 } } { \left[ \frac { { t }^{ 3 } }{ 3 } \right] }_{ 0 }^{ \tau }=\frac { { I }_{ 0 }^{ 2 } }{ { 3 } }\)
\( \left[ \because t=\tau \right] \)
\(\Rightarrow { I }_{ rms }=\sqrt { \bar { { I }^{ 2 } } } =\sqrt { \left( \frac { { I }_{ 0 }^{ 2 } }{ 3 } \right) } =\frac { { I }_{ 0 } }{ \sqrt { 3 } } \)

\({ I }_{ rms }=\sqrt { \frac { { I }_{ 1 }^{ 2 }+{ I }_{ 2 }^{ 2 }+{ I }_{ 3 }^{ 2 } }{ 3 } } =\sqrt { \frac { { I }^{ 2 }+{ -2 }^{ 2 }+{ 1 }^{ 2 } }{ 3 } }\)
\(=\sqrt { \frac { 6 }{ 3 } } =\sqrt { 2 } =1.414\) 

When the switch is thrown from the OFF position (open circuit) to the ON position (closed circuit), then neither B nor A and the angle between B and A does not change.Thus, no change in magnetic flux linked with coil occur, hence no electromotive force is produced and consequently, no current will flow in the circuit.

As the train is intercepting vertical components of the earth's magnetic field.The flux linked with the changes.So, there will be an induced emf.