1
JEE Main 2019 (Online) 8th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
The magnetic field of an electromagnetic wave is given by :-

$$\mathop B\limits^ \to = 1.6 \times {10^{ - 6}}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {2\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right){{Wb} \over {{m^2}}}$$

The associated electric field will be :-
A
$$\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z - 6 \times {{10}^{15}}t} \right)\left( -2{\mathop i\limits^ \wedge + \mathop {j}\limits^ \wedge } \right){V \over m}$$
B
$$\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z - 6 \times {{10}^{15}}t} \right)\left( 2{\mathop i\limits^ \wedge + \mathop {j}\limits^ \wedge } \right){V \over m}$$
C
$$\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {\mathop i\limits^ \wedge - \mathop {2j}\limits^ \wedge } \right){V \over m}$$
D
$$\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( -{\mathop i\limits^ \wedge + \mathop {2j}\limits^ \wedge } \right){V \over m}$$
2
JEE Main 2019 (Online) 8th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Two magnetic dipoles X and Y are placed at a separation d, with their axes perpendicular to each other. The dipole moment of Y is twice that of X. A particle of charge q is passing, through their midpoint P, at angle q = 45° with the horizontal line, as shown in figure. What would be the magnitude of force on the particle at that instant ?
(d is much larger than the dimensions of the dipole) JEE Main 2019 (Online) 8th April Evening Slot Physics - Magnetics Question 208 English
A
$$ \left( {{{{\mu _0}} \over {4\pi }}} \right){2M \over {{{\left( {d/2} \right)}^3}}} \times qv$$
B
$$ \left( {{{{\mu _0}} \over {4\pi }}} \right){M \over {{{\left( {d/2} \right)}^3}}} \times qv$$
C
$$\sqrt 2 \left( {{{{\mu _0}} \over {4\pi }}} \right){M \over {{{\left( {d/2} \right)}^3}}} \times qv$$
D
0
3
JEE Main 2019 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
A circular coil having N turns and radius r carries a current I. It is held in the XZ plane in a magnetic field B$${\mathop i\limits^ \wedge }$$ . The torque on the coil due to the magnetic field is :
A
$${{B{r^2}I} \over {\pi N}}$$
B
B$$\pi $$r2IN
C
Zero
D
$${{B\pi{r^2}I} \over { N}}$$
4
JEE Main 2019 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
A thin strip 10 cm long is on a U shaped wire of negligible resistance and it is connected to a spring of spring constant 0.5 Nm–1 (see figure). The assembly is kept in a uniform magnetic field of 0.1 T. If the strip is pulled from its equilibrium position and released, the number of oscillation it performs before its amplitude decreases by a factor of e is N. If the mass of the strip is 50 grams, its resistance 10W and air drag negligible, N will be close to : JEE Main 2019 (Online) 8th April Morning Slot Physics - Magnetics Question 212 English
A
50000
B
1000
C
5000
D
10000
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