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The strength of the magnetic field at a point r near a long straight current carrying wire is B. The field at a distance r/2 will be
An infinitely long straight conductor is bent into the shape as shown in the figure. It carries a current of i ampere and the radius of the circular loop is rmetre. Then the magnetic induction at its centre will be
The given shape is equivalent to the following diagram
The field at due to straight part of conductor isThe field at due to straight part of conductor is
The field at due to circular coil is Both fields will act in the opposite direction, hence the total field at O.
The magnetic moment of a current carrying loop is 2.1 × 10-25 amp × m2 . The magnetic field at a point on its axis at a distance of is
Field at a point from the centre of a current carrying loop on the axis is
The earth's magnetic field at a certain place has a horizontal component 0.3 Gauss and the total strength 0.5 Gauss. The angle of dip is
The horizontal component of the earth's magnetic field is 0.22 Gauss and total magnetic field is 0.4 Gauss. The angle of dip. is
By using BH = Bcos ϕ
When a ferromagnetic material is heated to temperature above its Curie temperature, the material
When a ferromagnetic material in heated above its curie temperature then it behaves like paramagnetic material.
In hydrogen atom, an electron is revolving in the orbit of radius 0.53 Å with 6.6 × 1015 rotations/second. Magnetic field produced at the centre of the orbit is
Four charges A, B, C, D are shown here, moving with equal speeds in a uniform magnetic field. Which charge experiences a maximum force?
As we know that F = qvB sin θ
Where, θ between the velocity and the magnetic field. Greater angle leads to maximum force. In given charges, angle is greater in case of (C), between magnetic field and velocity.
A Metal Rod Pq (Carrying Current From P To Q) Is Placed Perpendicular To The Infinite Long Wire Carrying Current I0. If This Arrangement Lies In A Horizontal Plane; Then The Rod Pq Will Rotate:
If given arrangement lies in a horizontal plane; then the rod PQ will rotate clockwise due to torque.
A coil having number of turns N and cross-sectional area A is rotated in a uniform magnetic field B with an angular velocity ω. The maximum value of the emf induced in it is
The flux linking with the coil at any instant t is given as Φ = NBA cos ωt
Therefore, the maximum value of emf is
Emax = NBAω
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