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Electromagnetic Waves question

2019 · 8 Apr · Shift 2 · Q51
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  5. /2019 · 8 Apr · Shift 2 · Q51

Electromagnetic Waves question

2019 · 8 Apr · Shift 2 · Q51

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
The magnetic field of an electromagnetic wave is given by :- B→=1.6×10−6cos⁡(2×107z+6×1015t)(2i∧+j∧)Wbm2\mathop B\limits^ \to = 1.6 \times {10^{ - 6}}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {2\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right){{Wb} \over {{m^2}}}B→​=1.6×10−6cos(2×107z+6×1015t)(2i∧​+j∧​)m2Wb​ The associated electric field will be :-
  1. A
    E→=4.8×102cos⁡(2×107z−6×1015t)(−2i∧+j∧)Vm\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z - 6 \times {{10}^{15}}t} \right)\left( -2{\mathop i\limits^ \wedge + \mathop {j}\limits^ \wedge } \right){V \over m}E→​=4.8×102cos(2×107z−6×1015t)(−2i∧​+j∧​)mV​
  2. B
    E→=4.8×102cos⁡(2×107z−6×1015t)(2i∧+j∧)Vm\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z - 6 \times {{10}^{15}}t} \right)\left( 2{\mathop i\limits^ \wedge + \mathop {j}\limits^ \wedge } \right){V \over m}E→​=4.8×102cos(2×107z−6×1015t)(2i∧​+j∧​)mV​
  3. C
    E→=4.8×102cos⁡(2×107z+6×1015t)(i∧−2j∧)Vm\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {\mathop i\limits^ \wedge - \mathop {2j}\limits^ \wedge } \right){V \over m}E→​=4.8×102cos(2×107z+6×1015t)(i∧​−2j∧​)mV​
  4. D
    E→=4.8×102cos⁡(2×107z+6×1015t)(−i∧+2j∧)Vm\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( -{\mathop i\limits^ \wedge + \mathop {2j}\limits^ \wedge } \right){V \over m}E→​=4.8×102cos(2×107z+6×1015t)(−i∧​+2j∧​)mV​
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