Explanation
The work done in turning a magnet of magnetic moment $$M$$ by an angle of $$90^{o}$$ from the meridian is $$n$$ times the corresponding work done to turn it through an angle of $$60^{o}$$ from the meridian, where $$n$$ is given by :
A magnet of moment $$4 \ Am^{2}$$ is suspended in a uniform magnetic field of induction $$5\times 10^{-4} \ T$$ with its axis at right angles. When the magnet is released from it's position, the kinetic energy that it gains in passing to the stable equilibrium position is :
A magnet of moment $$1.2\ Am^{2}$$ is kept suspended in a magnetic field of induction $$2\times 10^{-6}$$. The work done in rotating it through 120$$^{o}$$ is:
Assertion: When radius of a circular wire carrying current is doubled, its magnetic moment becomes four times
Reason: Magnetic moment is directly proportional to area of the loop
The magnetic moment of a bar magnet is 0.256 amp.m$$^{2}$$. Its pole strength is 400 milli amp. m. It is cut into two equal pieces and these two pieces are arranged at right angles to each other with their unlike poles in contact (or like poles in contact). The resultant magnetic moment of the system is
A magnet is in stable equilibrium in a uniform magnetic field. It is deflected by $$60^{o}$$ and the workdone is equal to $$W$$. In deflecting the magnet further by $$30^{o}$$, Work done is:
$$B=\dfrac{{{\mu }_{o}}}{4\pi }\dfrac{I}{a}\left( \sin {{\phi }_{1}}+\sin {{\phi }_{2}} \right)$$
Where angles $${{\phi }_{1}}$$ and $${{\phi }_{2}}$$ are as shown in the figure.
Correct Option is D.
Explanation for the correct answer:
Step 1: Note all the given values.
The magnetic field produced by a current-carrying straing conductor carrying current $$I$$ at a point P at a distance $$r$$ from the wire is calculated as follows:
In the given problem, we must calculate the Magnetic Field at a distance of $$3m$$ from one end of the wire, which is lying on the wire's axis.
Therefore, $${{\phi }_{1}}=0\,and\,{{\phi }_{2}}=180$$
Step 2: Calculate Field at given point.
Magnetic Field at a point at distance $$3m$$ from one end of the wire lying on the axis of the wire is,
$$B = \dfrac{{{\mu _o}}}{{4\pi }}\dfrac{I}{a}\left( {\sin 0 + \sin 180} \right)$$
$$ \Rightarrow B = \dfrac{{{\mu _o}}}{{4\pi }}\dfrac{I}{a}\left( {0 + 0} \right)$$
$$ \Rightarrow B = 0$$
As a result, the net Magnetic Field at a position $$3m$$ away from one end of the wire lying on the wire's axis is zero.
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