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

4.23 A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,

(a) the wire intersects the axis,

(b) the wire is turned from N-S to northeast-northwest direction,

(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

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Payal Gupta | Contributor-Level 10

5 months ago

4.23 Magnetic field strength, B = 1.5 T

Radius of the cylindrical region, r = 10 cm = 0.1 m

Current in the wire passing through the cylindrical region, I = 7 A

If the wire intersect the axis, then the length of the wire is the diameter of the cylindrical region, then l = 2r = 0.2 m

Angle between the magnetic field, $θ = 90 °$

Magnetic force acting on the wire is given by the relation,

F = BIl $\sin ? θ$ = 1.5 $\times 7 \times 0.2 \times \sin ? 90 °$ = 2.1 N

Hence, a force of 2.1 N acts on the wire in a vertically downward direction.

If the wire is turned from N-S to NE-NW direction, new length of the wire can be given as

$l_{1} = \frac{l}{\sin ? θ}$

Angle between magnetic field and current $θ$ = 45 $°$

Force on the wire,

F = BI $l_{1} \sin ? θ$ =

...more

4.23 Magnetic field strength, B = 1.5 T

Radius of the cylindrical region, r = 10 cm = 0.1 m

Current in the wire passing through the cylindrical region, I = 7 A

If the wire intersect the axis, then the length of the wire is the diameter of the cylindrical region, then l = 2r = 0.2 m

Angle between the magnetic field, $θ = 90 °$

Magnetic force acting on the wire is given by the relation,

F = BIl $\sin ? θ$ = 1.5 $\times 7 \times 0.2 \times \sin ? 90 °$ = 2.1 N

Hence, a force of 2.1 N acts on the wire in a vertically downward direction.

If the wire is turned from N-S to NE-NW direction, new length of the wire can be given as

$l_{1} = \frac{l}{\sin ? θ}$

Angle between magnetic field and current $θ$ = 45 $°$

Force on the wire,

F = BI $l_{1} \sin ? θ$ = BIl = 1.5 $\times 7 \times$ 0.2 = 2.1 N

A vertically downward force of 2.1 N acts on the wire.

When the wire is lowered from the axis by distance, d = 6.0 cm = 0.06 m

Let $l_{2}$ be the new length and $l_{2}$ is given by

${(\frac{l_{2}}{2})}^{2}$ = 4 (d + r) = 4(6 + 10) = 64

$l_{2}$ =16 cm = 0.16 m

$M a g n e t i c f o r c e$ $F_{2}$ = BI $l_{2}$ = 1.5 $\times 7 \times 0.16$ = 1.68 N

The force acts vertically downwards.

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<div>4.23 Magnetic field strength, B = 1.5 T<div>Radius of the cylindrical region, r = 10 cm = 0.1 mCurrent in the wire passing through the cylindrical region, I = 7 AIf the wire intersect the axis, then the length of the wire is the diameter of the cylindrical region, then l = 2r = 0.2 mAngle between the magnetic field, <math> <mi>θ</mi> <mo>=</mo> <mn>90</mn> <mo>°</mo> </math> Magnetic force acting on the wire is given by the relation,F = BIl <math> <mrow> <mrow> <mi mathvariant="normal">sin</mi> </mrow> <mo>?</mo> <mrow> <mi>θ</mi> </mrow> </mrow> </math> = 1.5 <math> <mo>×</mo> <mn>7</mn> <mo>×</mo> <mn>0.2</mn> <mo>×</mo> <mrow> <mrow> <mi mathvariant="normal">sin</mi> </mrow> <mo>?</mo> <mrow> <mn>90</mn> <mo>°</mo> </mrow> </mrow> </math> = 2.1 NHence, a force of 2.1 N acts on the wire in a vertically downward direction.If the wire is turned from N-S to NE-NW direction, new length of the wire can be given as <math> <msub> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mo>=</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mrow> <mrow> <mi mathvariant="normal">sin</mi> </mrow> <mo>?</mo> <mrow> <mi>θ</mi> </mrow> </mrow> </mrow> </mrow> </mfrac> </math> Angle between magnetic field and current <math> <mi>θ</mi> </math> = 45 <math> <mo>°</mo> </math> Force on the wire,F = BI <math> <msub> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mrow> <mrow> <mi mathvariant="normal">sin</mi> </mrow> <mo>?</mo> <mrow> <mi>θ</mi> </mrow> </mrow> </math> = BIl = 1.5 <math> <mo>×</mo> <mn>7</mn> <mo>×</mo> </math> 0.2 = 2.1 NA vertically downward force of 2.1 N acts on the wire.When the wire is lowered from the axis by distance, d = 6.0 cm = 0.06 mLet <math> <msub> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> be the new length and <math> <msub> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> is given by <math> <msup> <mrow> <mrow> <mo>(</mo> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> = 4 (d + r) = 4(6 + 10) = 64</div><div> <math> <msub> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> =16 cm = 0.16 m <math> <mi mathvariant="normal">M</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">g</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">c</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">f</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">r</mi> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">e</mi> </math> <math> <msub> <mrow> <mrow> <mi>F</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> = BI <math> <msub> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> = 1.5 <math> <mo>×</mo> <mn>7</mn> <mo>×</mo> <mn>0.16</mn> </math> = 1.68 NThe force acts vertically downwards.</div></div>

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