JEE Main 20198 Apr, Evening shiftPhysicsMediumElectromagnetic Waves120× since 2002The 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 :-AOption 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∧)mVBOption 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∧)mVCOption 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∧)mVDOption 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∧)mVCheckSkip · show solution
