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  • For the solenoid shown in the fig. the magnetic field at point P will be <img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAMQAAAB4CAIAAAAizMoyAAAZyUlEQVR4nO2deXgb5Z3HBwp092npdp8AgW

For the solenoid shown in the fig. the magnetic field at point P will be

Option: 1

\frac{\mu_o n i}{4}(\sqrt{3}+1)


Option: 2

\frac{\sqrt{3}}{4} \mu_o n I


Option: 3

\frac{\mu oni}{2} (\sqrt{3}+1)


Option: 4

\frac{\mu_o n i}{4}(\sqrt{3}-1)


Answers (1)

As we learned

The magnetic field in finite length solenoid -

B=\frac{\mu_o}{4\pi }(2\pi ni)(Sin\alpha +Sin\beta ),n=\frac{N}{l}

N=total number of turns 

 B=\frac{\mu_o}{4\pi}(2\pi ni)[sin\alpha +sin\beta ]

From figure:

\alpha = (90-30)=60^{\circ}

\beta = (90-60)=30^{\circ}

B=\frac{\mu_oni}{2}[sin60+sin30]=\frac{\mu_oni}{4}[\sqrt{3}+1]

 

Posted by

Sumit Saini

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