{"id":87105,"date":"2025-06-01T04:29:06","date_gmt":"2025-06-01T04:29:06","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87105"},"modified":"2025-06-01T04:29:06","modified_gmt":"2025-06-01T04:29:06","slug":"at-a-given-point-in-space-the-electric-field-associated-with-an-elect","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/at-a-given-point-in-space-the-electric-field-associated-with-an-elect\/","title":{"rendered":"At a given point in space, the electric field associated with an elect"},"content":{"rendered":"

At a given point in space, the electric field associated with an electromagnetic wave is given by $\\vec{E} = (2\\hat{i} – 1.5\\hat{j})e^{i[k_0(3x+4y)-\\omega t]}$. At the same point, which one among the following is the correct value of the unit vector $(\\hat{B})$ of the magnetic field associated with this electromagnetic wave ?<\/p>\n

[amp_mcq option1=”$-\\hat{k}$” option2=”$1.5\\hat{i} + 2\\hat{j}$” option3=”$1.5\\hat{i} – 2\\hat{j}$” option4=”$3\\hat{i} – 4\\hat{j}$” correct=”option1″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC Geoscientist – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct value for the unit vector of the magnetic field is $-\\hat{k}$.
\n<\/section>\n
\nFor a plane electromagnetic wave propagating in a dielectric medium, the electric field vector ($\\vec{E}$), the magnetic field vector ($\\vec{B}$), and the wave vector ($\\vec{k}$) are mutually perpendicular. The direction of wave propagation is given by the direction of $\\vec{E} \\times \\vec{B}$, which is the same as the direction of $\\vec{k}$.
\nThe given electric field is $\\vec{E} = (2\\hat{i} – 1.5\\hat{j})e^{i[k_0(3x+4y)-\\omega t]}$. The term $k_0(3x+4y)$ represents $\\vec{k} \\cdot \\vec{r}$. This indicates that the wave vector $\\vec{k}$ is in the direction $3\\hat{i} + 4\\hat{j}$.
\nThe direction of $\\vec{E}$ is given by the amplitude vector $(2\\hat{i} – 1.5\\hat{j})$. Let this be $\\vec{E}_0$.
\nWe need to find a unit vector $\\hat{B}$ such that $\\vec{E}_0 \\times \\hat{B}$ is in the direction of $3\\hat{i} + 4\\hat{j}$.
\nLet’s check option A: $\\hat{B} = -\\hat{k}$.
\n$\\vec{E}_0 \\times \\hat{B} = (2\\hat{i} – 1.5\\hat{j}) \\times (-\\hat{k}) = (2\\hat{i} \\times -\\hat{k}) + (-1.5\\hat{j} \\times -\\hat{k}) = 2\\hat{j} + 1.5\\hat{i} = 1.5\\hat{i} + 2\\hat{j}$.
\nThe direction of this vector $1.5\\hat{i} + 2\\hat{j}$ is indeed the same as $3\\hat{i} + 4\\hat{j}$ (since $1.5\\hat{i} + 2\\hat{j} = \\frac{1}{2}(3\\hat{i} + 4\\hat{j}) \\times 2 = 3\\hat{i} + 4\\hat{j}$).
\nThus, with $\\hat{B} = -\\hat{k}$, $\\vec{E}_0 \\times \\hat{B}$ is in the correct direction of wave propagation.
\nAlso, check perpendicularity: $\\vec{E}_0 \\cdot \\hat{B} = (2\\hat{i} – 1.5\\hat{j}) \\cdot (-\\hat{k}) = 0$ and $(3\\hat{i} + 4\\hat{j}) \\cdot (-\\hat{k}) = 0$, confirming $\\vec{E}$ and $\\vec{k}$ are perpendicular to $\\vec{B}$.
\nOptions B, C, and D are not unit vectors, and therefore cannot be the unit vector $\\hat{B}$.
\n<\/section>\n
\nThe ratio of the magnitudes of the electric and magnetic fields in vacuum is constant, $|E|\/|B| = c$, where c is the speed of light. In a medium, $|E|\/|B| = v$, where v is the speed of light in the medium. The directions are related by $\\vec{E} \\times \\vec{B} = v \\mu \\epsilon \\vec{E} \\times \\vec{E} = v \\vec{k}$ (assuming $\\vec{E}$ and $\\vec{B}$ are in phase). More accurately, the direction of propagation is $\\vec{E} \\times \\vec{B}$.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

At a given point in space, the electric field associated with an electromagnetic wave is given by $\\vec{E} = (2\\hat{i} – 1.5\\hat{j})e^{i[k_0(3x+4y)-\\omega t]}$. At the same point, which one among the following is the correct value of the unit vector $(\\hat{B})$ of the magnetic field associated with this electromagnetic wave ? [amp_mcq option1=”$-\\hat{k}$” option2=”$1.5\\hat{i} + … <\/p>\n

Detailed SolutionAt a given point in space, the electric field associated with an elect<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1091],"tags":[1103,1128,1274],"class_list":["post-87105","post","type-post","status-publish","format-standard","hentry","category-upsc-geoscientist","tag-1103","tag-physics","tag-wave-motion","no-featured-image-padding"],"yoast_head":"\nAt a given point in space, the electric field associated with an elect<\/title>\n<meta name=\"description\" content=\"The correct value for the unit vector of the magnetic field is $-hat{k}$. For a plane electromagnetic wave propagating in a dielectric medium, the electric field vector ($vec{E}$), the magnetic field vector ($vec{B}$), and the wave vector ($vec{k}$) are mutually perpendicular. The direction of wave propagation is given by the direction of $vec{E} times vec{B}$, which is the same as the direction of $vec{k}$. The given electric field is $vec{E} = (2hat{i} - 1.5hat{j})e^{i[k_0(3x+4y)-omega t]}$. The term $k_0(3x+4y)$ represents $vec{k} cdot vec{r}$. This indicates that the wave vector $vec{k}$ is in the direction $3hat{i} + 4hat{j}$. The direction of $vec{E}$ is given by the amplitude vector $(2hat{i} - 1.5hat{j})$. Let this be $vec{E}_0$. We need to find a unit vector $hat{B}$ such that $vec{E}_0 times hat{B}$ is in the direction of $3hat{i} + 4hat{j}$. Let's check option A: $hat{B} = -hat{k}$. $vec{E}_0 times hat{B} = (2hat{i} - 1.5hat{j}) times (-hat{k}) = (2hat{i} times -hat{k}) + (-1.5hat{j} times -hat{k}) = 2hat{j} + 1.5hat{i} = 1.5hat{i} + 2hat{j}$. The direction of this vector $1.5hat{i} + 2hat{j}$ is indeed the same as $3hat{i} + 4hat{j}$ (since $1.5hat{i} + 2hat{j} = frac{1}{2}(3hat{i} + 4hat{j}) times 2 = 3hat{i} + 4hat{j}$). Thus, with $hat{B} = -hat{k}$, $vec{E}_0 times hat{B}$ is in the correct direction of wave propagation. Also, check perpendicularity: $vec{E}_0 cdot hat{B} = (2hat{i} - 1.5hat{j}) cdot (-hat{k}) = 0$ and $(3hat{i} + 4hat{j}) cdot (-hat{k}) = 0$, confirming $vec{E}$ and $vec{k}$ are perpendicular to $vec{B}$. Options B, C, and D are not unit vectors, and therefore cannot be the unit vector $hat{B}$.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/at-a-given-point-in-space-the-electric-field-associated-with-an-elect\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"At a given point in space, the electric field associated with an elect\" \/>\n<meta property=\"og:description\" content=\"The correct value for the unit vector of the magnetic field is $-hat{k}$. For a plane electromagnetic wave propagating in a dielectric medium, the electric field vector ($vec{E}$), the magnetic field vector ($vec{B}$), and the wave vector ($vec{k}$) are mutually perpendicular. The direction of wave propagation is given by the direction of $vec{E} times vec{B}$, which is the same as the direction of $vec{k}$. The given electric field is $vec{E} = (2hat{i} - 1.5hat{j})e^{i[k_0(3x+4y)-omega t]}$. The term $k_0(3x+4y)$ represents $vec{k} cdot vec{r}$. This indicates that the wave vector $vec{k}$ is in the direction $3hat{i} + 4hat{j}$. The direction of $vec{E}$ is given by the amplitude vector $(2hat{i} - 1.5hat{j})$. Let this be $vec{E}_0$. We need to find a unit vector $hat{B}$ such that $vec{E}_0 times hat{B}$ is in the direction of $3hat{i} + 4hat{j}$. Let's check option A: $hat{B} = -hat{k}$. $vec{E}_0 times hat{B} = (2hat{i} - 1.5hat{j}) times (-hat{k}) = (2hat{i} times -hat{k}) + (-1.5hat{j} times -hat{k}) = 2hat{j} + 1.5hat{i} = 1.5hat{i} + 2hat{j}$. The direction of this vector $1.5hat{i} + 2hat{j}$ is indeed the same as $3hat{i} + 4hat{j}$ (since $1.5hat{i} + 2hat{j} = frac{1}{2}(3hat{i} + 4hat{j}) times 2 = 3hat{i} + 4hat{j}$). Thus, with $hat{B} = -hat{k}$, $vec{E}_0 times hat{B}$ is in the correct direction of wave propagation. Also, check perpendicularity: $vec{E}_0 cdot hat{B} = (2hat{i} - 1.5hat{j}) cdot (-hat{k}) = 0$ and $(3hat{i} + 4hat{j}) cdot (-hat{k}) = 0$, confirming $vec{E}$ and $vec{k}$ are perpendicular to $vec{B}$. Options B, C, and D are not unit vectors, and therefore cannot be the unit vector $hat{B}$.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/at-a-given-point-in-space-the-electric-field-associated-with-an-elect\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T04:29:06+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"2 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"At a given point in space, the electric field associated with an elect","description":"The correct value for the unit vector of the magnetic field is $-hat{k}$. For a plane electromagnetic wave propagating in a dielectric medium, the electric field vector ($vec{E}$), the magnetic field vector ($vec{B}$), and the wave vector ($vec{k}$) are mutually perpendicular. The direction of wave propagation is given by the direction of $vec{E} times vec{B}$, which is the same as the direction of $vec{k}$. The given electric field is $vec{E} = (2hat{i} - 1.5hat{j})e^{i[k_0(3x+4y)-omega t]}$. The term $k_0(3x+4y)$ represents $vec{k} cdot vec{r}$. This indicates that the wave vector $vec{k}$ is in the direction $3hat{i} + 4hat{j}$. The direction of $vec{E}$ is given by the amplitude vector $(2hat{i} - 1.5hat{j})$. Let this be $vec{E}_0$. We need to find a unit vector $hat{B}$ such that $vec{E}_0 times hat{B}$ is in the direction of $3hat{i} + 4hat{j}$. Let's check option A: $hat{B} = -hat{k}$. $vec{E}_0 times hat{B} = (2hat{i} - 1.5hat{j}) times (-hat{k}) = (2hat{i} times -hat{k}) + (-1.5hat{j} times -hat{k}) = 2hat{j} + 1.5hat{i} = 1.5hat{i} + 2hat{j}$. The direction of this vector $1.5hat{i} + 2hat{j}$ is indeed the same as $3hat{i} + 4hat{j}$ (since $1.5hat{i} + 2hat{j} = frac{1}{2}(3hat{i} + 4hat{j}) times 2 = 3hat{i} + 4hat{j}$). Thus, with $hat{B} = -hat{k}$, $vec{E}_0 times hat{B}$ is in the correct direction of wave propagation. Also, check perpendicularity: $vec{E}_0 cdot hat{B} = (2hat{i} - 1.5hat{j}) cdot (-hat{k}) = 0$ and $(3hat{i} + 4hat{j}) cdot (-hat{k}) = 0$, confirming $vec{E}$ and $vec{k}$ are perpendicular to $vec{B}$. Options B, C, and D are not unit vectors, and therefore cannot be the unit vector $hat{B}$.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/at-a-given-point-in-space-the-electric-field-associated-with-an-elect\/","og_locale":"en_US","og_type":"article","og_title":"At a given point in space, the electric field associated with an elect","og_description":"The correct value for the unit vector of the magnetic field is $-hat{k}$. For a plane electromagnetic wave propagating in a dielectric medium, the electric field vector ($vec{E}$), the magnetic field vector ($vec{B}$), and the wave vector ($vec{k}$) are mutually perpendicular. The direction of wave propagation is given by the direction of $vec{E} times vec{B}$, which is the same as the direction of $vec{k}$. The given electric field is $vec{E} = (2hat{i} - 1.5hat{j})e^{i[k_0(3x+4y)-omega t]}$. The term $k_0(3x+4y)$ represents $vec{k} cdot vec{r}$. This indicates that the wave vector $vec{k}$ is in the direction $3hat{i} + 4hat{j}$. The direction of $vec{E}$ is given by the amplitude vector $(2hat{i} - 1.5hat{j})$. Let this be $vec{E}_0$. We need to find a unit vector $hat{B}$ such that $vec{E}_0 times hat{B}$ is in the direction of $3hat{i} + 4hat{j}$. Let's check option A: $hat{B} = -hat{k}$. $vec{E}_0 times hat{B} = (2hat{i} - 1.5hat{j}) times (-hat{k}) = (2hat{i} times -hat{k}) + (-1.5hat{j} times -hat{k}) = 2hat{j} + 1.5hat{i} = 1.5hat{i} + 2hat{j}$. The direction of this vector $1.5hat{i} + 2hat{j}$ is indeed the same as $3hat{i} + 4hat{j}$ (since $1.5hat{i} + 2hat{j} = frac{1}{2}(3hat{i} + 4hat{j}) times 2 = 3hat{i} + 4hat{j}$). Thus, with $hat{B} = -hat{k}$, $vec{E}_0 times hat{B}$ is in the correct direction of wave propagation. Also, check perpendicularity: $vec{E}_0 cdot hat{B} = (2hat{i} - 1.5hat{j}) cdot (-hat{k}) = 0$ and $(3hat{i} + 4hat{j}) cdot (-hat{k}) = 0$, confirming $vec{E}$ and $vec{k}$ are perpendicular to $vec{B}$. Options B, C, and D are not unit vectors, and therefore cannot be the unit vector $hat{B}$.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/at-a-given-point-in-space-the-electric-field-associated-with-an-elect\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T04:29:06+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"2 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/at-a-given-point-in-space-the-electric-field-associated-with-an-elect\/","url":"https:\/\/exam.pscnotes.com\/mcq\/at-a-given-point-in-space-the-electric-field-associated-with-an-elect\/","name":"At a given point in space, the electric field associated with an elect","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T04:29:06+00:00","dateModified":"2025-06-01T04:29:06+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct value for the unit vector of the magnetic field is $-\\hat{k}$. For a plane electromagnetic wave propagating in a dielectric medium, the electric field vector ($\\vec{E}$), the magnetic field vector ($\\vec{B}$), and the wave vector ($\\vec{k}$) are mutually perpendicular. The direction of wave propagation is given by the direction of $\\vec{E} \\times \\vec{B}$, which is the same as the direction of $\\vec{k}$. The given electric field is $\\vec{E} = (2\\hat{i} - 1.5\\hat{j})e^{i[k_0(3x+4y)-\\omega t]}$. The term $k_0(3x+4y)$ represents $\\vec{k} \\cdot \\vec{r}$. This indicates that the wave vector $\\vec{k}$ is in the direction $3\\hat{i} + 4\\hat{j}$. The direction of $\\vec{E}$ is given by the amplitude vector $(2\\hat{i} - 1.5\\hat{j})$. Let this be $\\vec{E}_0$. We need to find a unit vector $\\hat{B}$ such that $\\vec{E}_0 \\times \\hat{B}$ is in the direction of $3\\hat{i} + 4\\hat{j}$. Let's check option A: $\\hat{B} = -\\hat{k}$. $\\vec{E}_0 \\times \\hat{B} = (2\\hat{i} - 1.5\\hat{j}) \\times (-\\hat{k}) = (2\\hat{i} \\times -\\hat{k}) + (-1.5\\hat{j} \\times -\\hat{k}) = 2\\hat{j} + 1.5\\hat{i} = 1.5\\hat{i} + 2\\hat{j}$. The direction of this vector $1.5\\hat{i} + 2\\hat{j}$ is indeed the same as $3\\hat{i} + 4\\hat{j}$ (since $1.5\\hat{i} + 2\\hat{j} = \\frac{1}{2}(3\\hat{i} + 4\\hat{j}) \\times 2 = 3\\hat{i} + 4\\hat{j}$). Thus, with $\\hat{B} = -\\hat{k}$, $\\vec{E}_0 \\times \\hat{B}$ is in the correct direction of wave propagation. Also, check perpendicularity: $\\vec{E}_0 \\cdot \\hat{B} = (2\\hat{i} - 1.5\\hat{j}) \\cdot (-\\hat{k}) = 0$ and $(3\\hat{i} + 4\\hat{j}) \\cdot (-\\hat{k}) = 0$, confirming $\\vec{E}$ and $\\vec{k}$ are perpendicular to $\\vec{B}$. Options B, C, and D are not unit vectors, and therefore cannot be the unit vector $\\hat{B}$.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/at-a-given-point-in-space-the-electric-field-associated-with-an-elect\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/at-a-given-point-in-space-the-electric-field-associated-with-an-elect\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/at-a-given-point-in-space-the-electric-field-associated-with-an-elect\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC Geoscientist","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-geoscientist\/"},{"@type":"ListItem","position":3,"name":"At a given point in space, the electric field associated with an elect"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87105","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=87105"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87105\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=87105"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=87105"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=87105"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}