The magnetic field of an electromagnetic wave is given by:

\({\rm{\vec B}} = 1.6 \times {10^{ - 6}}{\rm{cos}}\left( {2 \times {{10}^7}{\rm{z}} + 6 \times {{10}^{15}}{\rm{t}}} \right)\left( {2\hat i + \hat j} \right)\frac{{{\rm{Wb}}}}{{{{\rm{m}}^2}}}\)

The associated electric field will be:This question was previously asked in

JEE Mains Previous Paper 1 (Held On: 08 Apr 2019 Shift 2)

Option 4 : \({\rm{\vec E}} = 4.8 \times {10^2}{\rm{cos}}\left( {2 \times {{10}^7}{\rm{z}} + 6 \times {{10}^{15}}{\rm{t}}} \right)\left( {\hat i - 2\hat j} \right)\frac{{\rm{V}}}{{\rm{m}}}\)

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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90 Questions
360 Marks
180 Mins

**Concept:**

**Electromagnetic waves:**

Electromagnetic waves are formed when an electric field comes in contact with a magnetic field. They are hence known as ‘electromagnetic’ waves. The electric field and magnetic field of an electromagnetic wave are perpendicular (at right angles) to each other.

**Calculation:**

The magnetic field is given by the formula:

B(x, t) = B_{max} cos (kz – ωt) ----(1)

\(\vec B = 1.6 \times {10^{ - 6}}{\rm{cos}}\left( {2 \times {{10}^7}{\rm{z}} + 6 \times {{10}^{15}}{\rm{t}}} \right)\left( {2\hat i + {\rm{\hat j}}} \right)\frac{{{\rm{Wb}}}}{{{{\rm{m}}^2}}}\) ----(2)

On comparing equation (1) and (2),

K = 2 × 10^{7}

ω = 6 × 10^{15 }

B_{max }= 1.6 × 10^{-6}

The electric field is given by the formula:

E(x, t) = E_{max} cos (kz – ωt) ----(3)

Now, substituting the compared values in equation (3)

\({\rm{E}}\left( {{\rm{x}},{\rm{t}}} \right) = {{\rm{E}}_{{\rm{max}}}}\cos \left( {2 \times {{10}^7}{\rm{z}} - 6 \times {{10}^{15}}{\rm{t}}} \right)\left( {{\rm{x\hat i}} + {\rm{y\hat j}}} \right)\) ----(4)

Now, the value of maximum electric field is:

E_{max }= cB_{max}

Where, c = Speed of light = 3 × 10^{8} m/s

On substituting the values,

⇒ E_{max} = 3 × 10^{8 }× 1.6 × 10^{-6}

∴ E_{max }= 4.8 × 10^{2} T

Now, we need to substitute the above value in equation 4,

\({\rm{E}}\left( {{\rm{x}},{\rm{t}}} \right) = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}{\rm{z}} - 6 \times {{10}^{15}}{\rm{t}}} \right)\left( {{\rm{x}}\hat i + {\rm{y\hat j}}} \right)\) ----(5)

Now, we need to determine the x and y value in equation (5).

Since, the electric field and the magnetic field is perpendicular to each other. The dot product of both the vector is zero.

\( \Rightarrow \left( {2\hat i + {\rm{\hat j}}} \right).\left( {{\rm{\;x\hat i}} + {\rm{y\hat j}}} \right) = 0\)

⇒ 2x + y = 0 ----(6)

Any set of two numbers that satisfies the above equation are the values of x and y.

Let x = 1,

⇒ 2(1) + y = 0

∴ y = -2

On substituting the x and y in equation (6), the equation is getting satisfied.

Now, we need to substitute the value of x and y in equation (5).

\(\therefore {\rm{E}}\left( {{\rm{x}},{\rm{t}}} \right) = {\rm{\vec E}} = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}{\rm{z}} - 6 \times {{10}^{15}}{\rm{t}}} \right)\left( {{\rm{\hat i}} - 2{\rm{\hat j}}} \right)\frac{{\rm{V}}}{{\rm{m}}}\)