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A closely wound solenoid of 2000 turns and area of cross-section 1.5\times 10^{-4}m^{2} carries a current of 2.0 A. It suspended through its centre and prependicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5\times 10^{-2} tesla making an angle of 30^{o} with the axis of the solenoid. The torque on the solenoid will be

  • Option 1)

    3\times 10^{-2}N-m

  • Option 2)

    3\times 10^{-3}N-m

  • Option 3)

    1.5\times 10^{-3}N-m

  • Option 4)

    1.5\times 10^{-2}N-m

 

Answers (1)

best_answer

 

Torque -

underset{T}{
ightarrow}=underset{M}{
ightarrow}	imes underset{B}{
ightarrow}=M=NiA

T=MBsin 	heta =NBiAsin 	heta

 

- wherein

M - magnetic moment 

 

 \left | \vec{T} \right |=\left | NI \vec{A}\vec{B} \right |=NIAB\sin \Theta

2000\times 2\times 1.5\times 10^{-4}\times 5\times 10^{-2}\times \sin 30^{\circ}

1.5\times 10^{-2} N-m


Option 1)

3\times 10^{-2}N-m

Incorrect Option

Option 2)

3\times 10^{-3}N-m

Incorrect Option

Option 3)

1.5\times 10^{-3}N-m

Incorrect Option

Option 4)

1.5\times 10^{-2}N-m

Correct Option

Posted by

Aadil

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