1
JEE Main 2020 (Online) 7th January Morning Slot
+4
-1
If the magnetic field in a plane electromagnetic wave is given by
$$\overrightarrow B$$ = 3 $$\times$$ 10-8 sin(1.6 $$\times$$ 103x + 48 $$\times$$ 1010t)$$\widehat j$$ T, then what will be expression for electric field ?
A
$$\overrightarrow E$$ = (9sin(1.6 $$\times$$ 103x + 48 $$\times$$ 1010t)$$\widehat k$$ V/m)
B
$$\overrightarrow E$$ = (60sin(1.6 $$\times$$ 103x + 48 $$\times$$ 1010t)$$\widehat k$$ V/m)
C
$$\overrightarrow E$$ = (3 $$\times$$ 10-8 sin(1.6 $$\times$$ 103x + 48 $$\times$$ 1010t)$$\widehat i$$ V/m)
D
$$\overrightarrow E$$ = (3 $$\times$$ 10-8 sin(1.6 $$\times$$ 103x + 48 $$\times$$ 1010t)$$\widehat j$$ V/m)
2
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
3
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
4
JEE Main 2019 (Online) 12th April Morning Slot
+4
-1
A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of 40 $$\pi$$ rad s–1 about its axis, perpendicular to its plane. If the magnetic field at its centre is 3.8 × 10–9 T, then the charge carried by the ring is close to ($$\mu$$0 = 4$$\pi$$ × 10–7 N/A2 ).
A
7 × 10–6 C
B
4 × 10–5 C
C
2 × 10–6 C
D
3 × 10–5 C
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