JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 6)
The magnetic field of a plane electromagnetic
wave is given by :
$$$\overline B = {B_0}\widehat i\left[ {\cos (kz - \omega t)} \right] + {B_i}\widehat j\cos (kz + \omega t)$$$ B0 = 3 × 10–5 T and B1 = 2 × 10–6 T.
The rms value of the force experienced by a stationary charge Q = 10–4 C at z = 0 is closest to :
$$$\overline B = {B_0}\widehat i\left[ {\cos (kz - \omega t)} \right] + {B_i}\widehat j\cos (kz + \omega t)$$$ B0 = 3 × 10–5 T and B1 = 2 × 10–6 T.
The rms value of the force experienced by a stationary charge Q = 10–4 C at z = 0 is closest to :
0.6 N
0.9 N
3 × 10–2 N
0.1 N
Explanation
Maximum electric field E = (B) (C)
$$\overrightarrow {{E_0}} = \left( {3 \times {{10}^{ - 5}}} \right)c\left( { - \widehat j} \right)$$
$$\overrightarrow {{E_1}} = \left( {2 \times {{10}^{ - 6}}} \right)c\left( { - \widehat i} \right)$$
Maximum force
$${\overrightarrow F _{net}} = {10^{ - 4}} \times 3 \times {10^8}\sqrt {{{\left( {3 \times {{10}^{ - 5}}} \right)}^2} + {{\left( {2 \times {{10}^{ - 6}}} \right)}^2}} = 0.9N$$
$${F_{rms}} = {{{F_0}} \over {\sqrt 2 }} = 0.6\,N$$ (approx)
$$\overrightarrow {{E_0}} = \left( {3 \times {{10}^{ - 5}}} \right)c\left( { - \widehat j} \right)$$
$$\overrightarrow {{E_1}} = \left( {2 \times {{10}^{ - 6}}} \right)c\left( { - \widehat i} \right)$$
Maximum force
$${\overrightarrow F _{net}} = {10^{ - 4}} \times 3 \times {10^8}\sqrt {{{\left( {3 \times {{10}^{ - 5}}} \right)}^2} + {{\left( {2 \times {{10}^{ - 6}}} \right)}^2}} = 0.9N$$
$${F_{rms}} = {{{F_0}} \over {\sqrt 2 }} = 0.6\,N$$ (approx)
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