The magnetic field vector of an electromagnetic wave is given by B = B₀ = $\frac{\hat{i} + \hat{j}}{\sqrt[]{2}} c o s (k z - ω t);$ where $\hat{i}, \hat{j}$ represents unit vector along x and y-axis respectively. At t = 0 s, two electric charges q1 of 4π coulomb and q2 and 2p coulomb located at $(0, 0, \frac{π}{k}) a n d (0, 0, \frac{3 π}{k}),$ respectively, have the same velocity of 0.5 c $\hat{i}$ (where c is the velocity of light). The ratio of the force acting on charge q1 to q2 is:

Option 1 -

$1 : \sqrt[]{2}$

Option 2 -

$\sqrt[]{2} : 1$

Option 3 -

2 : 1

Option 4 -

$2 \sqrt[]{2} : 1$

4 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

Force on q₁ :

$F_{q_{1}} = q (V \times B) = 4 π (0.5 c \hat{i}) \times \frac{B_{0}}{\sqrt[]{2}} (c o s (k z - ω t) \hat{i} + c o s (k z - w t) \hat{j})$

$= 4 π (0.5 c) (\frac{B_{0}}{\sqrt[]{2}}) c o s (k \frac{π}{k}) \hat{k}$

For on q₂:

= $2 π (0.5 c) (\frac{B_{0}}{\sqrt[]{2}}) c o s (k \frac{3 π}{k}) \hat{k}$

So ratio will be 2 : 1

Force on q1 : <math> <mrow> <msub> <mrow> <mi>F</mi> </mrow> <mrow> <msub> <mrow> <mi>q</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> </msub> <mo>=</mo> <mi>q</mi> <mrow> <mo> (</mo> <mrow> <mi>V</mi> <mo>×</mo> <mi>B</mi> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>4</mn> <mi>π</mi> <mrow> <mo> (</mo> <mrow> <mn>0</mn> <mo>.</mo> <mn>5</mn> <mi>c</mi> <mover accent="true"> <mi>i</mi> <mo>^</mo> </mover> </mrow> <mo>)</mo> </mrow> <mo>×</mo> <mfrac> <mrow> <msub> <mrow> <mi>B</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> <mrow> <mo> (</mo> <mrow> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mrow> <mo> (</mo> <mrow> <mi>k</mi> <mi>z</mi> <mo>−</mo> <mi>ω</mi> <mi>t</mi> </mrow> <mo>)</mo> </mrow> <mover accent="true"> <mi>i</mi> <mo>^</mo> </mover> <mo>+</mo> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mrow> <mo> (</mo> <mrow> <mi>k</mi> <mi>z</mi> <mo>−</mo> <mi>w</mi> <mi>t</mi> </mrow> <mo>)</mo> </mrow> <mover accent="true"> <mi>j</mi> <mo>^</mo> </mover> </mrow> <mo>)</mo> </mrow> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1815997275"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mo>=</mo> <mn>4</mn> <mi>π</mi> <mrow> <mo> (</mo> <mrow> <mn>0</mn> <mo>.</mo> <mn>5</mn> <mi>c</mi> </mrow> <mo>)</mo> </mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>B</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mrow> <mo> (</mo> <mrow> <mi>k</mi> <mfrac> <mrow> <mi>π</mi> </mrow> <mrow> <mi>k</mi> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mover accent="true"> <mi>k</mi> <mo>^</mo> </mover> </mrow> </math>      <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1815997276"> </o:OLEObject></xml><! [endif]->For on q2:= <math> <mrow> <mn>2</mn> <mi>π</mi> <mrow> <mo> (</mo> <mrow> <mn>0</mn> <mo>.</mo> <mn>5</mn> <mi>c</mi> </mrow> <mo>)</mo> </mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>B</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mrow> <mo> (</mo> <mrow> <mi>k</mi> <mfrac> <mrow> <mn>3</mn> <mi>π</mi> </mrow> <mrow> <mi>k</mi> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mover accent="true"> <mi>k</mi> <mo>^</mo> </mover> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1815997277"> </o:OLEObject></xml><! [endif]->So ratio will be 2 : 1

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