Magnetic effect of current was discovered by
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- 1A deutron of kinetic energy $50\, keV$ is describing a circular orbit of radius $0.5$ $metre$ in a plane perpendicular to magnetic field $\overrightarrow B $. The kinetic energy of the proton that describes a circular orbit of radius $0.5$ $metre$ in the same plane with the same $\overrightarrow B $ is........$keV$View Solution
- 2A charged particle of mass $m$ and charge $q$ travels on a circular path of radius $r$ that is perpendicular to a magnetic field $B$. The time taken by the particle to complete one revolution isView Solution
- 3A long solenoid with $20$ $turns$ per $cm$ is made. To produce a magnetic field of $20$ $millitesla$ inside the solenoid, the necessary current will nearly be......$ampere$View Solution
- 4A galvanometer has a resistance of $50\ \Omega$ and it allows maximum current of $5 \mathrm{~mA}$. It can be converted into voltmeter to measure upto $100 \mathrm{~V}$ by connecting in series a resistor of resistanceView Solution
- 5A galvanometer of $10 \,\Omega$ resistance gives full scale deflection with $0.01$ ampere of current. It is to be converted into an ammeter for measuring $10$ ampere current. The value of shunt resistance required will beView Solution
- 6A proton (mass $ = 1.67 \times {10^{ - 27}}\,kg$ and charge $ = 1.6 \times {10^{ - 19}}\,C)$ enters perpendicular to a magnetic field of intensity $2$ $weber/{m^2}$ with a velocity $3.4 \times {10^7}\,m/\sec $. The acceleration of the proton should beView Solution
- 7The earth’s magnetic field at a given point is $0.5 \times {10^{ - 5}}\,Wb{\rm{ - }}{m^{ - 2}}$. This field is to be annulled by magnetic induction at the center of a circular conducting loop of radius $5.0\,cm$. The current required to be flown in the loop is nearly......$A$View Solution
- 8A particle with ${10^{ - 11}}\,coulomb$ of charge and ${10^{ - 7}}\,kg$ mass is moving with a velocity of ${10^8}\,m/s$ along the $y$-axis. A uniform static magnetic field $B = 0.5\,Tesla$ is acting along the $x$-direction. The force on the particle isView Solution
- 9View SolutionFrom Ampere's circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is :
- 10The figure shows a conducting loop $ABCDA$ placed in a uniform magnetic field perpendicular to its plane. The part $ABC$ is the $(3/4)^{th}$ portion of the square of side length $l$ . The part $ADC$ is a circular arc of radius $R$ . The points $A$ and $C$ are connected to a battery which supply a current $I$ to the circuit. The magnetic force on the loop due to the field $B$ isView Solution
