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  • A small flat search coils of area with 20 closely wound turns is positioned normal

A small flat search coils of area \mathrm{4cm}^{2} with 20 closely wound turns is positioned normal to the field direction and then quickly snatched out of the field region. The total charge flowing in the coil (measured by a ballistic galvanometer connected to the coil) is 7.5 mC. The resistance of the coil and galvanometer is \mathrm{0.8\Omega } . The field strength of the magnet is:

Option: 1

1.25 T


Option: 2

1.50 T


Option: 3

0.75 T


Option: 4

2.10 T


Answers (1)

best_answer

\text { Here, } A=4 \mathrm{~cm}^2=4 \times 10^{-4} \mathrm{~m}^2, \mathrm{~N}=20

\text { Final flux }, \phi_{\mathrm{f}}=0                  (when the coil is removed from the field)

\mathrm{q}=7.5 \mathrm{mC}=7.5 \times 10^{-3} \mathrm{C}, \mathrm{R}=0.8 \Omega

\mathrm{As \; I=\frac{\varepsilon}{R}=\frac{-N(d \phi / d t)}{R} }

\mathrm{\mathrm{Idt}=-\frac{\mathrm{N}}{\mathrm{R}} \mathrm{d} \phi }

\mathrm{\text { Also charge, } q=\int \mathrm{Idt}=\int_{\phi_{\mathrm{i}}}^{\phi_{\mathrm{f}}}-\frac{\mathrm{N}}{\mathrm{R}} \mathrm{d} \phi=-\frac{\mathrm{N}}{\mathrm{R}}\left(\phi_{\mathrm{f}}-\phi_{\mathrm{i}}\right)=\frac{\mathrm{N}}{\mathrm{R}}\left(\phi_{\mathrm{i}}-\phi_{\mathrm{f}}\right)}

\mathrm{\mathrm{q}=\frac{\mathrm{N}}{\mathrm{R}} \phi_{\mathrm{i}}}

Now, initial flux per turn when coil is normal to the field,\mathrm{\phi_{\mathrm{i}}=\mathrm{BA}}

\mathrm{\therefore q=\frac{N B A}{R} \quad \text { or } \quad B=\frac{q R}{N A}=\frac{7.5 \times 10^{-3} \times 0.8}{20 \times 4 \times 10^{-4}}=0.75 \mathrm{~T}}

 

 

 

Posted by

Riya

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