Practice free Magnetic Effects of Current and Magnetism (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
A solenoid of 1.5 meter length and 4 cm diameter possesses 10 turn per cm. A current of 5 Amp is flowing through it. The magnetci induction at axis inside the solenoid is
$ n = 10 turns/ m = 1000 turns /m \Rightarrow B = \mu_o nI = 2 \pi \times 10^{-3} Tesla $
A conducting circular loop of radius a carries a constant current I. It is placed in a uniformmagnetic field $ \vec B$ , such that $\vec B $ is perpendicular to the plane of the Loop. The magnetic force acting on the Loop is ...................
Net force on a current carrying closed Loop is always zero if it is placed in a uniform mag. field.
A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a permanent magnet such that $ \vec B $ is in plane of the coil. If due to a current I in the triangle a torque $ \tau $ acts on it, the side l of the triangle is
$ In \triangle CAD $ $ AC^2 = AD^2 + DC^2 $ $ l^2 = { l^2 \over 4 } + DC^2 $ $ DC = { \sqrt 3 \over 2 } l $ $ Area of \triangle ABC $ $ A = {1 \over 2 } (l) \left( {\sqrt 2 \over 2 }l \right) $ $ A = { 1 \over 4 } \sqrt 3 l^2 $ $ torques acting on \triangle ABC is $ $ \tau = IAB sin \theta $ $ = I \left( {1 \over 4 } \sqrt 3 l^2 \right) B sin ^\circ $ $ \theta = 90 ^\circ $ $ \tau = { \sqrt 3 \over 4 } I l^2 B $ $ \therefore l^2 = - { 4 \tau \over \sqrt 3 I B } $ $ \therefore l^2 = { 4 \tau \over \sqrt 3 I B } $ $ \therefore l = 2 \left ({ \tau \over \sqrt I B } \right) ^{1/2 } $
A coil having N turns is wound tightly in the formof a spiral with inner and outer radii "a" and "b" respectively. When a current I passes through the coil, the magnetic field at the centre is ...............
$ No. of turns per unit width = { N \over b-a} $ $ \therefore the no.of turns in thickness dx is dN = \left( N \over b-a\right) dx $ $ \therefore mag. field at the centre is dB = dN \left( \mu_o I \over 2x \right) $ $ dB = \left( {N \over b-a} \right) { \mu_o I \over 2x} .dx $ $ \therefore mag.field B = \int dB$ $ = { \mu_o NI \over 2 (b-a) } n \left( { n \over a } \right) $
An iron rod of length L and magnetic moment M is bent in the form of a semi circle. Now its magnetic moment will be
On bending a rod its pole strangth remains unchanged where as its magnetic moment changes. M' = m (2 R ) $ = m \left( 2 { L \over \pi } \right) = { 2 \over \pi } mL = { 2m \over \pi } $
Magnetic intensity for an axial point due to a short bar magnet of magnetic moment M is given by
A magnet of magnetic moment M and pole strength m is divided in two equal parts, then magnetic moment of each part will be
Case (i)
If cut along the axis of magnet of length l , then new pole strength m' = m/2
and new length l ' = l
$ \therefore New magnetic moment M' = { m \over 2 } \times l = { ml \over 2 } = { M \over 2 } $
Case (ii)
If cut perpendicular to the axis of magnet, then new pole strength m' = m
and new length l' = l /2
$ \therefore New magnetic moment M' = { m \over 2 } \times l = { ml \over 2 } = { M \over 2 } $
If a magnet of pole strengthm is divided into four parts such that the length and width of each part is half that of initial one, then the pole strength of each part will be
For each part m' = m/2
The magnetism of magnet is due to
The spin motion of electron
The magnetic field at a point x on the axis of a small bar magnet is equal to the at a point y on the equator of the same magnet. The ratio of the distances of x and y from the centre of the magnet is
$ on the axis B_1 = { 2m \over x^3 } $ $ on the equator B_2 = { M \over y^3 } $ $ As B_1 = B_2 $ $ { 2M \over x^3 } = { M \over y^3 } $ $ { x^3 \over y^3 } = 2 $ $ { x \over y } = 2 ^ { 1 \over 3 } $
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