Magnetic Field Due to a Straight Long Wire

Q. The direction of magnetic lines of forces close to a straight conductor carrying current will be

1. triangular
2. Radially outward
3. Circular in a plane perpendicular to the conductor
4. Helical

Q. If a copper rod carries a direct current, the magnetic field associated with the current will be

1. Only inside the rod
2. Only outside the rod
3. Both inside and outside the rod
4. Neither inside nor outside the rod

Q. a copper rod o length l is rotated about one end perpendicular to the uniorm magnetic ield B with constant angular velocity w . the induced em between its two ends is

Q.

When a long wire carrying a steady current is bent into a circular coil of one turn, the magnetic induction at its centre is B. When the same wire carrying the same current is bent to form a circular coil of n turns of a smaller radius, the magnetic induction at the centre will be

1. $\frac{B}{n}$

2. nB

3. $\frac{B}{n^2}$

4. $n^{2}B$

Q. A long vertical wire carrying a current of 1A in the upward direction is placed in a region where a horizontal magnetic field of $2.0\times {10}^{-4}T$ exists from south to north. Find the point where the resultant magnetic field is zero.

1. 1.0mm, to the west of the wire
2. 1.0 mm, to the East of the wire
3. 2.0 mmm to the North of the wire
4. 2.0 mm, to the South of the wire

Q. Two very thin metallic wires placed along X– and Y–axes carry equal currents as shown in figure. AB and CD are lines at ${45}^{\circ }$ with the axes with origin of axes at O. The magnetic fields will be zero on the line

1. AB
2. CD
3. Segment OB only of line AB
4. Segment OC only of line CD

Q. Work required to rotate given square frame by ${180}^{\circ }$ can be given as

1. $\frac{{\mu }_0}{2\pi }{i}_1{i}_{2}aln\left(2\right)$
2. $\frac{{\mu }_0}{\pi }{i}_1{i}_{2}aln\left(2\right)$
3. $\frac{{\mu }_0}{\pi }{i}_1{i}_{2}a$
4. $\frac{{\mu }_0}{2\pi }{i}_1{i}_{2}a$

Q.

Wires 1 and 2 carrying currents $i_1$ and $i_2$ respectively are inclined at an angle $\theta$ to each other. What is the force on a small element dl of wire 2 at a distance of r from wire 1 (as shown in figure) due to the magnetic field of wire1

1. $\frac{{\mu }_0}{2\pi r}{i}_1{i}_{2}dltan\theta$

2. $\frac{{\mu }_0}{2\pi r}{i}_1{i}_{2}dlsin\theta$

3. $\frac{{\mu }_0}{2\pi r}{i}_1{i}_{2}dlcos\theta$

4. $\frac{{\mu }_0}{4\pi r}{i}_1{i}_{2}dlsin\theta$

Q. Formula for calculating magnetic field due to infinite wire can be derived from the formula of magnetic field due to finite wire.

1. False
2. True

Q.

Wires 1 and 2 carrying currents $i_1$ and $i_2$ respectively are inclined at an angle $\theta$ to each other. What is the force on a small element dl of wire 2 at a distance of r from wire 1 (as shown in figure) due to the magnetic field of wire1

1. $\frac{{\mu }_0}{2\pi r}{i}_1{i}_{2}dltan\theta$

2. $\frac{{\mu }_0}{2\pi r}{i}_1{i}_{2}dlsin\theta$

3. $\frac{{\mu }_0}{2\pi r}{i}_1{i}_{2}dlcos\theta$

4. $\frac{{\mu }_0}{4\pi r}{i}_1{i}_{2}dlsin\theta$

Q. Find the magnetic field due to a square frame of side a carrying a current as shown at 0.

1. $Zero$
2. $\frac{\left(2{\mu }_{0}i\right)}{\pi a}$
3. $\frac{{\mu }_{0}i}{2\sqrt{2}\pi a}$
4. $2\frac{\sqrt{2}{\mu }_{0}i}{\pi a}$

Q.

Four infinitely ling wires are placed at the vertices of a square of side . Find B at its centre

1. $Zero$

2. $\frac{{\mu }_{0}i}{\pi a}$

3. $\frac{2{\mu }_{0}i}{\pi a}$

4. $\frac{4{\mu }_{0}i}{\pi a}$

Q.

A straight horizontal conducting rod of length 0.5 m and mass 50 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires. What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero? Ignore the mass of the wires and take g = 10 $m{s}^{-2}$.

1. 0.1 T

2. 0.2 T

3. 0.3 T

4. 0.4 T

Q. An infinitely long conductor PQR is bent to form a right angle as shown. A current I flows through PQR The magnetic field due to this current at the point M is $H_1$. Now another infinitely long straight conductor QS is connected at Q so that the current is I/2 in QR as well as in QS, The current in PQ remaining unchanged. The magnetic field at M is now $H_2$ The ratio $\frac{H_1}{H_2}$ is given by

1. $\frac{1}{2}$
2. $1$
3. $\frac{2}{3}$
4. $2$