1
JEE Main 2019 (Online) 8th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Two very long, straight, and insulated wires are kept at 90° angle from each other in xy-plane as shown in the figure. These wires carry currents of equal magnitude I, whose directions are shown in the figure. The net magnetic field at point P will be : JEE Main 2019 (Online) 8th April Evening Slot Physics - Magnetic Effect of Current Question 143 English
A
$${{ + {\mu _0}I} \over {\pi d}}\left( {\mathop z\limits^ \wedge } \right)$$
B
$$ - {{{\mu _0}I} \over {2\pi d}}\left( {\mathop x\limits^ \wedge + \mathop y\limits^ \wedge } \right)$$
C
Zero
D
$$ {{{\mu _0}I} \over {2\pi d}}\left( {\mathop x\limits^ \wedge + \mathop y\limits^ \wedge } \right)$$
2
JEE Main 2019 (Online) 8th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A circuit connected to an ac source of emf e = e0sin(100t) with t in seconds, gives a phase difference of $$\pi $$/4 between the emf e and current i. Which of the following circuits will exhibit this ?
A
RC circuit with R = 1 k$$\Omega $$ and C = 1μF
B
RL circuit with R = 1k$$\Omega $$ and L = 1mH
C
RC circuit with R = 1k$$\Omega $$ and C = 10 μF
D
RL circuit with R = 1 k$$\Omega $$ and L = 10 mH
3
JEE Main 2019 (Online) 8th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
$$\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( -{\mathop i\limits^ \wedge + \mathop {2j}\limits^ \wedge } \right){V \over m}$$
4
JEE Main 2019 (Online) 8th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to :-
A
0.2%
B
3.5%
C
0.7%
D
6.8%
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