For adjoining fig., The magnetic field at point, $P$ will be
A$\frac{\mu_0}{4 \pi} \odot$
B$\frac{\mu_0}{\pi} \otimes$
C$\frac{\mu_0}{2 \pi} \odot$
D$\frac{\mu_0}{2 \pi} \otimes$
AIPMT 2000, Medium
Download our app for free and get started
D$\frac{\mu_0}{2 \pi} \otimes$
d Magnetic field due to $5 A \rightarrow \frac{5 \mu_0}{2 \pi \times 2.5}=\frac{2 \mu_0}{2 \pi} \otimes$
Magnetic field due to $2.5 A \rightarrow \frac{2.5 \mu_0}{2 \pi \times 2.5}=\frac{\mu_0}{2 \pi} \odot$
Resultant Magnetic field $B_{net}=\frac{2 \mu_0}{2 \pi}-\frac{\mu_0}{2 \pi}=\frac{\mu_0}{2 \pi} \otimes$
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
A charged particle is released from rest in a region of steady uniform electric and magnetic fields which are parallel to each other the particle will move in a
A very long wire $ABDMNDC$ is shown in figure carrying current $I. AB$ and $BC$ parts are straight, long and at right angle. At $D$ wire forms a circular turn $DMND$ of radius $R. AB.$ $\mathrm{BC}$ parts are tangential to circular turn at $\mathrm{N}$ and $D$. Magnetic field at the centre of circle is
In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric potential $V$ and then made to describe semicircular paths of radius $R$ using a magnetic field $B$. If $V$ and $B$ are kept constant, the ratio $\left( {\frac{{{\text{charge on the ion}}}}{{{\text{mass of the ion}}}}} \right)$ will be proportional to
In an ionised sodium atom, an electron is moving in a circular path of radius $r$ with angular velocity $\omega $. The magnetic induction in $wb/m^2$ produced at the nucleus will be
A very long conducting wire is bent in a semicircular shape from $A$ to $B$ as shown in figure. The magnetic field at point $P$ for steady current configuration is given by:
An electron (charge $q$ $coulomb$) enters a magnetic field of $H$ $weber/{m^2}$ with a velocity of $v\,m/s$ in the same direction as that of the field the force on the electron is