1
JEE Main 2019 (Online) 12th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
2
JEE Main 2019 (Online) 10th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. if this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be:
A
$${m \over \pi }$$
B
$${{3m} \over \pi }$$
C
$${{2m} \over \pi }$$
D
$${{4m} \over \pi }$$
3
JEE Main 2019 (Online) 10th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is :
[Take $$\mu $$0 = 4$$\pi $$ × 10–7 NA–2]
A
3 $$\mu $$T
B
18 $$\mu $$T
C
9 $$\mu $$T
D
1 $$\mu $$T
4
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Two wires A & B are carrying currents I1 & I2 as shown in the figure. The separation between them is d. A third wire C carrying a current I is to be kept parallel to them at a distance x from A such that the net force acting on it is zero. The possible values of x are : JEE Main 2019 (Online) 10th April Morning Slot Physics - Magnetic Effect of Current Question 138 English
A
$$x = \left( {{{{I_1}} \over {{I_1} + {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} - {I_2}}}} \right)d$$
B
$$x = \left( {{{{I_2}} \over {{I_1} + {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} - {I_2}}}} \right)d$$
C
$$x = \left( {{{{I_1}} \over {{I_1} - {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} + {I_2}}}} \right)d$$
D
$$x = \pm {{{I_1}d} \over {{I_1} - {I_2}}}$$
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