1
JEE Main 2020 (Online) 8th January Morning Slot
+4
-1
Photon with kinetic energy of 1MeV moves from south to north. It gets an acceleration of 1012 m/s2 by an applied magnetic field (west to east). The value of magnetic field : (Rest mass of proton is 1.6 × 10–27 kg) :
A
0.71mT
B
7.1mT
C
0.071mT
D
71mT
2
JEE Main 2020 (Online) 7th January Evening Slot
+4
-1
A particle of mass m and charge q has an initial velocity $$\overrightarrow v = {v_0}\widehat j$$ . If an electric field $$\overrightarrow E = {E_0}\widehat i$$ and magnetic field $$\overrightarrow B = {B_0}\widehat i$$ act on the particle, its speed will double after a time:
A
$${{3m{v_0}} \over {q{E_0}}}$$
B
$${{\sqrt 2 m{v_0}} \over {q{E_0}}}$$
C
$${{\sqrt 3 m{v_0}} \over {q{E_0}}}$$
D
$${{2m{v_0}} \over {q{E_0}}}$$
3
JEE Main 2019 (Online) 12th April Evening Slot
+4
-1
An electron, moving along the x-axis with an initial energy of 100 eV, enters a region of magnetic field $$\overrightarrow B = \left( {1.5 \times {{10}^{ - 3}}T} \right)\widehat k$$ at S (See figure). The field extends between x = 0 and x = 2 cm. The electron is detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P and Q (on the screen) is : (electron’s charge = 1.6 × 10–19 C, mass of electron = 9.1 × 10–31 kg)
A
2.25 cm
B
12.87 cm
C
1.22 cm
D
11.65 cm
4
JEE Main 2019 (Online) 12th April Evening Slot
+4
-1
Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5A. (See figure) ($$\mu$$0 = 4$$\pi$$ × 10–7 N-A–2)
A
1.5 × 10–5 T
B
3.0 × 10–5 T
C
2.0 × 10–5 T
D
2.5 × 10–5 T
EXAM MAP
Medical
NEETAIIMS