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Standard XII

Physics

Exploiting Ampere's Law - Long Wire

A long soleno...

Question

A long solenoid is formed by winding $20 turns/cm$ . The current necessary to produce a magnetic field of $20 millitesla$ inside the solenoid will be approximately
$(\frac{μ_{0}}{4 π} = 10^{- 7} tesla-metre/ampere)$

8.0 A

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4.0 A

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2.0 A

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1.0 A

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Solution

The correct option is A $8.0 A$
$B = μ_{0} n i \Rightarrow i = \frac{B}{μ_{0} n} = \frac{20 \times 10^{- 3}}{4 π \times 10^{- 7} \times 20 \times 100}$
$= 7.9 amp \approx 8 amp$

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A long solenoid is formed by winding 20 turns/cm. The current necessary to produce a magnetic field of 20 millitesla inside the solenoid will be approximately (μ04π=10−7 tesla-metre/ampere)

The correct option is A 8.0 AB=μ0ni⇒i=Bμ0n=20×10−34π×10−7×20×100 =7.9 amp≈8 amp

A long solenoid is formed by winding $20 turns/cm$ . The current necessary to produce a magnetic field of $20 millitesla$ inside the solenoid will be approximately
$(\frac{μ_{0}}{4 π} = 10^{- 7} tesla-metre/ampere)$

The correct option is A $8.0 A$
$B = μ_{0} n i \Rightarrow i = \frac{B}{μ_{0} n} = \frac{20 \times 10^{- 3}}{4 π \times 10^{- 7} \times 20 \times 100}$
$= 7.9 amp \approx 8 amp$