Subjective Type
A closed coil having $$100$$ turns is rotated in a uniform magnetic field $$B=4.0\times { 10 }^{ -4 }T$$ about a diameter whcih is perpendicular to the field. The angular velocity of rotation is $$300$$ revolutions per minute. The area of the coil is $$25{cm}^{2}$$ and its resistance is $$4.0\Omega$$. Find (a) the average emf developed in half turn from a position where the coil is perpendicular to the magnetic field, (b) the average emf in a full turn and (c) the net charge displaced in part (a)
Solution
$$n=100turns,B=4\times { 10 }^{ -4 }T;A=25{ cm }^{ 2 }=25\times { 10 }^{ -4 }{ m }^{ 2 }$$
(a) when the coil is perpendicular to the field
$$\phi =nBA$$
when coil goes through half a turn
$$\phi =BA\cos { { 18 }^{ o } } =0-NBA$$
$$d\phi =2nBA$$
The coil undergoes 300rev, in 1 min
$$300\times 2\pi rad/min=10\pi rad/sec$$
$$10\pi $$ rad is swept in 1 sec
$$\pi /\pi $$ rad is swept $$1/10\pi \times \pi =1/10sec$$
$$\quad E=\cfrac { d\phi }{ { dt }^{ } } =\cfrac { 2nBA }{ dt } =\cfrac { 2\times 100\times 4\times { 10 }^{ -4 }\times 25\times { 10 }^{ -4 } }{ 1/10 } =2\times { 10 }^{ -3 }V\quad $$
(b) $${ \phi }_{ 1 }=nBA,{ \phi }_{ 2 }=nBA(\theta ={ 360 }^{ o })\quad ;d\phi =0\quad $$
(c) $$i=\cfrac { E }{ R } =\cfrac { 2\times { 10 }^{ -3 } }{ 4 } =\cfrac { 1 }{ 2 } \times { 10 }^{ -3 }=0.5\times { 10 }^{ -3 }=5\times { 10 }^{ -4 }\quad $$
$$q=idt=5\times { 10 }^{ -4 }\times 1/10=5\times { 10 }^{ -5 }C\quad $$
$$E=\cfrac { d\phi }{ { dt }^{ } } =\cfrac { 2nBA }{ dt } =\cfrac { 2\times 100\times 4\times { 10 }^{ -4 }\times 25\times { 10 }^{ -4 } }{ 1/10 } =2\times { 10 }^{ -3 }V$$
(b) $${ \phi }_{ 1 }=nBA,{ \phi }_{ 2 }=nBA(\theta ={ 360 }^{ o })\quad ;d\phi =0\quad $$
(c) $$i=\cfrac { E }{ R } =\cfrac { 2\times { 10 }^{ -3 } }{ 4 } =\cfrac { 1 }{ 2 } \times { 10 }^{ -3 }=0.5\times { 10 }^{ -3 }=5\times { 10 }^{ -4 }\quad $$
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