6.13 It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm² with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Ω. Estimate the field strength of magnet.

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Answered by

Payal Gupta | Contributor-Level 10

4 months ago

6.13 Area of the coil, A = 2 ${c m}^{2}$ = 2 $\times 10^{- 4}$ $m^{2}$

Number of turns, n = 25

Total charge flowing in the coil, Q = 7.5 mC = 7.5 $\times 10^{- 3}$ C

Total resistance of the coil and galvanometer, R = 0.50 Ω

We know, Induced current in the coil, I = $\frac{I n d u c e d e m f (e)}{R}$ ……… (1)

Induced emf is given by the equation, e = $-$ n $\frac{d \emptyset}{d t}$ ………………… (2),

Where d $\emptyset$ is the change of flux

Combining equations (1) and (2), we get

I = $\frac{- n \frac{d \emptyset}{d t}}{R}$ or Idt = $- \frac{n}{R}$ d $\emptyset$ ……… (3)

Initial flux through coil, $\emptyset_{i}$ = BA, where B = Magnetic field strengt

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6.13 Area of the coil, A = 2 ${c m}^{2}$ = 2 $\times 10^{- 4}$ $m^{2}$

Number of turns, n = 25

Total charge flowing in the coil, Q = 7.5 mC = 7.5 $\times 10^{- 3}$ C

Total resistance of the coil and galvanometer, R = 0.50 Ω

We know, Induced current in the coil, I = $\frac{I n d u c e d e m f (e)}{R}$ ……… (1)

Induced emf is given by the equation, e = $-$ n $\frac{d \emptyset}{d t}$ ………………… (2),

Where d $\emptyset$ is the change of flux

Combining equations (1) and (2), we get

I = $\frac{- n \frac{d \emptyset}{d t}}{R}$ or Idt = $- \frac{n}{R}$ d $\emptyset$ ……… (3)

Initial flux through coil, $\emptyset_{i}$ = BA, where B = Magnetic field strength

Final flux through coil, $\emptyset_{f}$ = 0

Integrating equation (3), we get

$\int I d t$ = $- \frac{n}{R} \int_{\emptyset_{i}}^{\emptyset_{f}} d \emptyset$

$\int I d t = Q = T o t a l c h a r g e f l o w i n g i n t h e c o i l$

Therefore

Q = $- \frac{n}{R} (\emptyset_{f} - \emptyset_{i}) = - \frac{n}{R} (0 - \emptyset_{i}) = \frac{n \emptyset_{i}}{R} = \frac{n B A}{R}$

B = $\frac{Q R}{n A}$ = $\frac{7.5 \times 10^{- 3} \times 0.5}{25 \times 2 \times 10^{- 4}}$ = 0.75 T

Hence, the field strength is 0.75 T

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