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Physics Ncert Solutions Class 12th Physics Class 12th Moving Charges and Magnetism

4.26 A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire? g = 9.8 m s^–2.

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Payal Gupta | Contributor-Level 10

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4.26 Length of the solenoid, L = 60 cm = 0.6 m

Radius of the solenoid, r = 4.0 cm = 0.04 m

It is given that there are 3 layers of windings of 300 turns each

Hence, total number of turns, n = 900

Length, l = 2 cm = 0.02 m

Mass of the wire, m = 2.5 g = 2.5 $\times 10^{- 3}$ kg $\times 10^{- 3}$

Current flowing through the wire, I = 6 A

Acceleration due to gravity, g = 9.8 m/ $s^{2}$

We know, magnetic field produced inside the solenoid, B = $\frac{μ_{0} n I}{L}$

where $μ_{0}$ = Permeability of free space = 4 $π \times 10^{- 7}$ T m $A^{- 1}$

Magnetic force is given by the relation

F = Bil = $\frac{μ_{0} n I}{L} \times i l$

Also the force on the wire is equal to the weight of the wire, F = mg

mg = $\frac{μ_{0} n I i l}{L}$

I = $\frac{m g L}{μ_{0} n i I}$ = $\frac{2.5 \times 10^{- 3} \times 9.8 \times 0.6}{4 π \times 10^{- 7} \times 900 \times 0.02 \times 6}$ = 108 A

Hence, the current flowing through the solenoid is 108 A

...more

4.26 Length of the solenoid, L = 60 cm = 0.6 m

Radius of the solenoid, r = 4.0 cm = 0.04 m

It is given that there are 3 layers of windings of 300 turns each

Hence, total number of turns, n = 900

Length, l = 2 cm = 0.02 m

Mass of the wire, m = 2.5 g = 2.5 $\times 10^{- 3}$ kg $\times 10^{- 3}$

Current flowing through the wire, I = 6 A

Acceleration due to gravity, g = 9.8 m/ $s^{2}$

We know, magnetic field produced inside the solenoid, B = $\frac{μ_{0} n I}{L}$

where $μ_{0}$ = Permeability of free space = 4 $π \times 10^{- 7}$ T m $A^{- 1}$

Magnetic force is given by the relation

F = Bil = $\frac{μ_{0} n I}{L} \times i l$

Also the force on the wire is equal to the weight of the wire, F = mg

mg = $\frac{μ_{0} n I i l}{L}$

I = $\frac{m g L}{μ_{0} n i I}$ = $\frac{2.5 \times 10^{- 3} \times 9.8 \times 0.6}{4 π \times 10^{- 7} \times 900 \times 0.02 \times 6}$ = 108 A

Hence, the current flowing through the solenoid is 108 A.

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4.26 Length of the solenoid, L = 60 cm = 0.6 mRadius of the solenoid, r = 4.0 cm = 0.04 mIt is given that there are 3 layers of windings of 300 turns eachHence, total number of turns, n = 900Length, l = 2 cm = 0.02 mMass of the wire, m = 2.5 g = 2.5 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> </msup> </math> kg <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> </msup> </math> <div>Current flowing through the wire, I = 6 AAcceleration due to gravity, g = 9.8 m/ <math> <msup> <mrow> <mrow> <mi>s</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> We know, magnetic field produced inside the solenoid, B = <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>μ</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> <mi>n</mi> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mi>L</mi> </mrow> </mrow> </mfrac> </math> where <math> <msub> <mrow> <mrow> <mi>μ</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math> = Permeability of free space = 4 <math> <mi>π</mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>7</mn> </mrow> </mrow> </msup> </math> T m <math> <msup> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </msup> </math> Magnetic force is given by the relationF = Bil = <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>μ</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> <mi>n</mi> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mi>L</mi> </mrow> </mrow> </mfrac> <mo>×</mo> <mi>i</mi> <mi>l</mi> </math> Also the force on the wire is equal to the weight of the wire, F = mgmg = <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>μ</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> <mi>n</mi> <mi>I</mi> <mi>i</mi> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mi>L</mi> </mrow> </mrow> </mfrac> </math> I = <math> <mfrac> <mrow> <mrow> <mi>m</mi> <mi>g</mi> <mi>L</mi> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>μ</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> <mi>n</mi> <mi>i</mi> <mi>I</mi> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mn>2.5</mn> <mi mathvariant="normal"> </mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>9.8</mn> <mo>×</mo> <mn>0.6</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> <mi>π</mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>7</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>900</mn> <mo>×</mo> <mn>0.02</mn> <mo>×</mo> <mn>6</mn> </mrow> </mrow> </mfrac> </math> = 108 A</div>Hence, the current flowing through the solenoid is 108 A.

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