A simple generator has a 300 loop square coil of side 20 cm turning in

1.	A simple generator has a 300 loop square coil of side 20 cm turning in a field of 0.7 T. How fast must it turn to produce a peak output of 210 V?
Answer» Given : Number of loops = 300 Length of the coil = 20 cm Magnetic field = 0.7 T EMF induced = 210 V ‎ ‎ To find : VELOCITY of the coil to produce a peak output of 210 V. ‎ ‎ Formula to be used : $\bf\:\maltese$ $\bf\color{black}{\:ε_{ind}~=~VBL}$ ‎where, $\sf\:ε_{ind}$ denotes emf induced L denotes length of the coil B denotes the magnetic field ‎ ‎ Solution : In the question it is mentioned that there are 300 loops so we NEED to modify the formula and multiply it with N which is equal to number of loops. ‎Therefore, $\sf\:ε_{ind}~=~N(VBL)$ Substituting the given values $\sf\longrightarrow\:210~=~300(V \times 0.7 \times 20)$ $\sf\longrightarrow\:210~=~300(V \times \frac{7}{10} \times 20)$ $\sf\longrightarrow\:210~=~300(V \times \frac{140}{10})$ $\sf\longrightarrow\:210~=~300(V \times \frac{14\cancel{0}}{1\cancel{0}})$ $\sf\longrightarrow\:210~=~300(V \times 14)$ $\sf\longrightarrow\:210~=~300 \times 14 \times V$ $\sf\longrightarrow\:\frac{210}{300 \times 14}~=~V$ $\sf\longrightarrow\:\frac{21\cancel{0}}{30\cancel{0} \times 14}~=~V$ $\sf\longrightarrow\:\frac{21}{30 \times 14}~=~V$ $\sf\longrightarrow\:\frac{\cancel{21}}{30 \times \cancel{14}}~=~V$ $\sf\longrightarrow\:\frac{3}{30 \times 2}~=~V$ $\sf\longrightarrow\:\frac{\cancel{3}}{\cancel{30} \times 2}~=~V$ $\sf\longrightarrow\:\frac{1}{10 \times 2}~=~V$ $\sf\longrightarrow\:\frac{1}{20}~=~V$ $\small{\underline{\boxed{\mathrm{\longrightarrow\:V~=~0.05~m/s}}}}$ ‎ ‎ Verification : LET's verify our result with the formula. $\sf\:ε_{ind}~=~N(VBL)$ $\sf\to\:210~=~300(0.05 \times 0.7 \times 20)$ $\sf\to\:210~=~300(\frac{5}{100} \times \frac{7}{10} \times 20)$ $\sf\to\:210~=~300(\frac{35}{1000} \times 20)$ $\sf\to\:210~=~300(\frac{700}{1000})$ $\sf\to\:210~=~300(\frac{7\cancel{00}}{10\cancel{00}})$ $\sf\to\:210~=~300(\frac{7}{10})$ $\sf\to\:210~=~\frac{2100}{10}$ $\sf\to\:210~=~\frac{210\cancel{0}}{1\cancel{0}}$ $\sf\to\:210~=~210$ Hence Verified !~ ═══════════════════════ Note : I have explained SIMPLE calculations as well therefore it's looking‎ long you can escape some of the steps . USING the formula it has got very short calculation ! :D ‎

A simple generator has a 300 loop square coil of side 20 cm turning in a field of 0.7 T. How fast must it turn to produce a peak output of 210 V?

Answer»

Given :

Number of loops = 300
Length of the coil = 20 cm
Magnetic field = 0.7 T
EMF induced = 210 V

‎

To find :

VELOCITY of the coil to produce a peak output of 210 V.

‎

Formula to be used :

$\bf\:\maltese$ $\bf\color{black}{\:ε_{ind}~=~VBL}$

‎where,

$\sf\:ε_{ind}$ denotes emf induced
L denotes length of the coil
B denotes the magnetic field

‎

Solution :

In the question it is mentioned that there are 300 loops so we NEED to modify the formula and multiply it with N which is equal to number of loops.

‎Therefore,

$\sf\:ε_{ind}~=~N(VBL)$

Substituting the given values

$\sf\longrightarrow\:210~=~300(V \times 0.7 \times 20)$

$\sf\longrightarrow\:210~=~300(V \times \frac{7}{10} \times 20)$

$\sf\longrightarrow\:210~=~300(V \times \frac{140}{10})$

$\sf\longrightarrow\:210~=~300(V \times \frac{14\cancel{0}}{1\cancel{0}})$

$\sf\longrightarrow\:210~=~300(V \times 14)$

$\sf\longrightarrow\:210~=~300 \times 14 \times V$

$\sf\longrightarrow\:\frac{210}{300 \times 14}~=~V$

$\sf\longrightarrow\:\frac{21\cancel{0}}{30\cancel{0} \times 14}~=~V$

$\sf\longrightarrow\:\frac{21}{30 \times 14}~=~V$

$\sf\longrightarrow\:\frac{\cancel{21}}{30 \times \cancel{14}}~=~V$

$\sf\longrightarrow\:\frac{3}{30 \times 2}~=~V$

$\sf\longrightarrow\:\frac{\cancel{3}}{\cancel{30} \times 2}~=~V$

$\sf\longrightarrow\:\frac{1}{10 \times 2}~=~V$

$\sf\longrightarrow\:\frac{1}{20}~=~V$

$\small{\underline{\boxed{\mathrm{\longrightarrow\:V~=~0.05~m/s}}}}$

‎

Verification :

LET's verify our result with the formula.

$\sf\:ε_{ind}~=~N(VBL)$

$\sf\to\:210~=~300(0.05 \times 0.7 \times 20)$

$\sf\to\:210~=~300(\frac{5}{100} \times \frac{7}{10} \times 20)$

$\sf\to\:210~=~300(\frac{35}{1000} \times 20)$

$\sf\to\:210~=~300(\frac{700}{1000})$

$\sf\to\:210~=~300(\frac{7\cancel{00}}{10\cancel{00}})$

$\sf\to\:210~=~300(\frac{7}{10})$

$\sf\to\:210~=~\frac{2100}{10}$

$\sf\to\:210~=~\frac{210\cancel{0}}{1\cancel{0}}$

$\sf\to\:210~=~210$

Hence Verified !~

═══════════════════════

Note : I have explained SIMPLE calculations as well therefore it's looking‎ long you can escape some of the steps . USING the formula it has got very short calculation !

‎

A simple generator has a 300 loop square coil of side 20 cm turning in a field of 0.7 T. How fast must it turn to produce a peak output of 210 V?

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A simple generator has a 300 loop square coil of side 20 cm turning in a field of 0.7 T. How fast must it turn to produce a peak output of 210 V?​

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A simple generator has a 300 loop square coil of side 20 cm turning in a field of 0.7 T. How fast must it turn to produce a peak output of 210 V?